From
$$ x+y+z=5 $$
$$ x=5-y-z $$
Substitute in second equation:
$$ 2(5-y-z)-y+z=9 $$
$$ 10-3y-z=9 $$
$$ 3y+z=1 \quad ...(1) $$
Substitute in third equation:
$$ (5-y-z)-2y+3z=16 $$
$$ 5-3y+2z=16 $$
$$ -3y+2z=11 \quad ...(2) $$
From (1):
$$ z=1-3y $$
Substitute in (2):
$$ -3y+2(1-3y)=11 $$
$$ -3y+2-6y=11 $$
$$ -9y=9 $$
$$ y=-1 $$
Then
$$ z=1-3(-1)=4 $$
Now
$$ x=5-(-1)-4 $$
$$ x=2 $$
Answer
$$ \boxed{(x,y,z)=(2,-1,4)} $$
$$ \boxed{\left(\frac12,\frac13,\frac14\right)} $$
$$ \boxed{(35,30,25)} $$
$$ $$ \boxed{\text{Infinitely many solutions}} $$
$$ $$ \boxed{\text{No solution}} $$
$$ $$ \boxed{\text{Unique solution}} $$
- Vani = $\boxed{24\text{ years}}$
- Father = $\boxed{51\text{ years}}$
- Grandfather = $\boxed{84\text{ years}}$
$$ \boxed{137} $$
Let the numbers of ₹5, ₹10 and ₹20 notes be 7, 3 and 2 respectively.
Check: 7×5 + 3×10 + 2×20 = 35 + 30 + 40 = ₹105. After interchanging first two counts: 3×5 + 7×10 + 2×20 = 15 + 70 + 40 = ₹125, increased by ₹20.
Answer: ₹5 notes = 7, ₹10 notes = 3, ₹20 notes = 2
Factor the first polynomial:
$$ x^4+3x^3-x-3 $$
Grouping:
$$ x^3(x+3)-1(x+3) $$
$$ =(x+3)(x^3-1) $$
$$ =(x+3)(x-1)(x^2+x+1) $$
Factor the second polynomial:
$$ x^3+x^2-5x+3 $$
Testing roots:
$$ x=1 $$
gives zero.
So,
$$ =(x-1)(x^2+2x-3) $$
$$ =(x-1)^2(x+3) $$
Common factors:
$$ (x-1)(x+3) $$
$$ =x^2+2x-3 $$
Answer
$$ \boxed{x^2+2x-3} $$
$$ x^4-1=(x^2-1)(x^2+1) $$
$$ =(x-1)(x+1)(x^2+1) $$
Factor second polynomial:
$$ x^3-11x^2+x-11 $$
Grouping:
$$ x^2(x-11)+1(x-11) $$
$$ =(x^2+1)(x-11) $$
Common factor:
$$ x^2+1 $$
Answer
$$ \boxed{x^2+1} $$
Factor first polynomial:
$$ 3x(x^3+2x^2-4x-8) $$
$$ =3x(x+2)(x^2-4) $$
$$ =3x(x+2)^2(x-2) $$
Factor second polynomial:
$$ 2x(2x^3+7x^2+4x-4) $$
$$ =2x(x+2)^2(2x-1) $$
Common factors:
$$ x(x+2)^2 $$
Answer
$$ \boxed{x(x^2+4x+4)} $$
Factor first polynomial:
$$ 3(x^3+x^2+x+1) $$
$$ =3(x+1)(x^2+1) $$
Factor second polynomial:
$$ 6(x^3+2x^2+x+2) $$
$$ =6(x+2)(x^2+1) $$
Common factor:
$$ 3(x^2+1) $$
Answer
$$ \boxed{3(x^2+1)} $$
Take: - LCM of coefficients = 8 - Highest powers: - $x^3$ - $y^2$
Answer
$$ \boxed{8x^3y^2} $$
LCM of coefficients:
$$ 36 $$
Highest powers: - $a^3$ - $b^2$ - $c$
Answer
$$ \boxed{36a^3b^2c} $$
LCM coefficient:
$$ 48 $$
Highest powers: - $m^2$ - $n^2$
Answer
$$ \boxed{48m^2n^2} $$
Factor:
$$ p^2-3p+2=(p-1)(p-2) $$
$$ p^2-4=(p-2)(p+2) $$
LCM:
$$ (p-1)(p-2)(p+2) $$
Answer
$$ \boxed{(p-1)(p-2)(p+2)} $$
Factor:
$$ 2x^2-5x-3=(2x+1)(x-3) $$
$$ 4x^2-36=4(x-3)(x+3) $$
LCM:
$$ 4(x+3)(2x+1)(x-3) $$
Answer
$$ \boxed{4(x+3)(2x+1)(x-3)} $$
Factor:
$$ 2x^2-3xy=x(2x-3y) $$
$$ 8x^3-27y^3=(2x-3y)(4x^2+6xy+9y^2) $$
LCM:
$$ x(2x-3y)(4x^2+6xy+9y^2) $$
Answer
$$ \boxed{x(2x-3y)(4x^2+6xy+9y^2)} $$
Answers Summary
- (i) GCD = x^2 + 2x - 3
- (ii) GCD = x^2 + 1
- (iii) GCD = x(x^2 + 4x + 4)
- (iv) GCD = 3(x^2 + 1)
| Question | Answer | |---|---| | 2(i) | $8x^3y^2$ | | 2(ii) | $-36a^3b^2c$ | | 2(iii) | $-48m^2n^2$ | | 2(iv) | $(p-1)(p-2)(p+2)$ | | 2(v) | $4(x+3)(2x+1)(x-3)$ | | 2(vi) | $2^3x^2(2x-3y)^3(4x^2+6xy+9y^2)$ |
$$ f(x)\times g(x)=\text{LCM}\times\text{GCD} $$
GCD
- GCD of coefficients:
$$ \gcd(21,35)=7 $$
- Lowest powers:
- $x^1$
- $y^1$
$$ \text{GCD}=7xy $$
LCM
- LCM of coefficients:
$$ \text{lcm}(21,35)=105 $$
- Highest powers:
- $x^2$
- $y^2$
$$ \text{LCM}=105x^2y^2 $$
Verification
$$ (21x^2y)(35xy^2) = 735x^3y^3 $$
$$ (105x^2y^2)(7xy) = 735x^3y^3 $$
Verified.
Answer
$$ \boxed{\text{LCM}=105x^2y^2,\quad \text{GCD}=7xy} $$
Factor:
$$ x^3-1=(x-1)(x^2+x+1) $$
$$ x^3+1=(x+1)(x^2-x+1) $$
So first polynomial becomes:
$$ (x-1)(x^2+x+1)(x+1) $$
Second polynomial:
$$ (x+1)(x^2-x+1) $$
GCD
There is no non-constant common factor, so
$$ \text{GCD}=1 $$
LCM
Take all factors:
$$ (x-1)(x+1)(x^2+x+1)(x^2-x+1) $$
Answer
$$ \boxed{ \text{LCM}=(x-1)(x+1)(x^2+x+1)(x^2-x+1) } $$
$$ \boxed{ \text{GCD}=1 } $$
Factor:
$$ x^2y+xy^2=xy(x+y) $$
$$ x^2+xy=x(x+y) $$
GCD
Common factor:
$$ x(x+y) $$
LCM
Take highest powers:
$$ xy(x+y) $$
Answer
$$ \boxed{\text{LCM}=xy(x+y)} $$
$$ \boxed{\text{GCD}=x(x+y)} $$
Factor:
$$ a^2+4a-12=(a+6)(a-2) $$
$$ a^2-5a+6=(a-2)(a-3) $$
LCM:
$$ (a+6)(a-2)(a-3) $$
Answer
$$ \boxed{(a+6)(a-2)(a-3)} $$
Factor:
$$ x^4-27a^3x = x(x^3-27a^3) $$
$$ =x(x-3a)(x^2+3ax+9a^2) $$
LCM:
$$ x(x-3a)^2(x^2+3ax+9a^2) $$
Answer
$$ \boxed{ x(x-3a)^2(x^2+3ax+9a^2) } $$
Using:
$$ f(x)\times g(x)=\text{LCM}\times\text{GCD} $$
After simplification,
$$ \text{GCD}=4x^2(x-1) $$
Answer
$$ \boxed{4x^2(x-1)} $$
Factor:
$$ x^3+y^3=(x+y)(x^2-xy+y^2) $$
Thus common factor:
$$ x^2-xy+y^2 $$
Answer
$$ \boxed{x^2-xy+y^2} $$
$$ \boxed{(a+2)(a-7)} $$
$$ \boxed{x^2+xy+y^2} $$
Answers Summary
| Question | LCM | GCD | |---|---|---| | 1(i) | $105x^2y^2$ | $7xy$ | | 1(ii) | $(x-1)(x+1)(x^2+x+1)(x^2-x+1)$ | $(x+1)$ | | 1(iii) | $xy(x+y)$ | $x(x+y)$ |
- (i) LCM = (a+6)(a-2)(a-3)
- (ii) LCM = x(x-3a)^2(x^2+3ax+9a^2)
- (i) GCD = 4x^2(x-1)
- (ii) GCD = x^2 - xy + y^2
- (i) Unknown polynomial = (a+2)(a-7)
- (ii) Unknown polynomial = x^2 + xy + y^2
> Note: > The original question images/formulae were incomplete in the provided source text. > Exact rational expressions were not fully visible in OCR extraction.
Solution Method
To reduce a rational expression to lowest form:
1. Factor numerator completely. 2. Factor denominator completely. 3. Cancel common factors. 4. Mention excluded values separately.
Example Format
$$ \frac{x^2-9}{x^2-3x} $$
Factor:
$$ =\frac{(x-3)(x+3)}{x(x-3)} $$
Cancel common factor:
$$ =\frac{x+3}{x} $$
with excluded value
$$ x\ne3,\ 0 $$
> Answers could not be reconstructed fully because the original expressions were not visible in the provided content.
UNIT 3 : Algebra
Factor p(x): x^2 - 5x - 14 = (x-7)(x+2).
Given (x-7)(x+2)/q(x) = (x-7)/(x+2) ⇒ q(x) = (x+2)^2 = x^2 + 4x + 4.
Answer: q(x) = x^2 + 4x + 4
> Note: > The original rational expression was incomplete in the provided OCR/source text. > Full numerator and denominator were not visible.
General Method
If
$$ A-B=C $$
then
$$ B=A-C $$
So, the rational expression to be subtracted can be found by:
1. Taking LCM of denominators 2. Simplifying 3. Subtracting appropriately
Pari’s one hour work:
$$ \frac14 $$
Yuvan’s one hour work:
$$ \frac16 $$
Together:
$$ \frac14+\frac16 $$
LCM of 4 and 6 is 12.
$$ =\frac{3}{12}+\frac{2}{12} $$
$$ =\frac{5}{12} $$
Together they complete:
$$ \frac{5}{12} $$
of the work in one hour.
Hence total time required:
$$ =\frac{12}{5} $$
hours.
$$ =2\frac25 $$
hours.
Answer
$$ \boxed{\frac{12}{5}\text{ hours}} $$
or
$$ \boxed{2\frac25\text{ hours}} $$
Let the price of 1 kg banana = ₹x, then price of 1 kg apple = ₹2x.
Apples
Amount spent = ₹1800 → quantity = 1800 / (2x) = 900 / x kg.
Bananas
Amount spent = ₹600 → quantity = 600 / x kg.
Total quantity: (900/x) + (600/x) = 50 ⇒ 1500 / x = 50 ⇒ x = 30.
Thus banana price = ₹30/kg, apple price = ₹60/kg.
Quantities:
Apples = 1800 / 60 = 30 kg.
Bananas = 600 / 30 = 20 kg.
Answer: 30 kg apples and 20 kg bananas.
> Note: > The rational expressions in Question 1 were incomplete in the OCR/source text provided. > Exact expressions were not visible.
General Method
To find square roots by factorization:
1. Factor numerator completely. 2. Factor denominator completely. 3. Pair equal factors. 4. Take one factor from each pair.
Example:
$$ \sqrt{\frac{a^2b^4}{c^6}} = \frac{ab^2}{c^3} $$
Recognize perfect square form:
$$ a^2+2ab+b^2 $$
where
$$ a=2x,\quad b=5 $$
Thus,
$$ 4x^2+20x+25=(2x+5)^2 $$
Taking square root:
$$ \sqrt{4x^2+20x+25}=2x+5 $$
Answer
$$ \boxed{2x+5} $$
Rearrange:
$$ 9x^2+16y^2+25z^2-24xy+30xz-40yz $$
Recognize:
$$ (3x-4y+5z)^2 $$
since
$$ (3x)^2=9x^2 $$
$$ (-4y)^2=16y^2 $$
$$ (5z)^2=25z^2 $$
Middle terms:
$$ 2(3x)(-4y)=-24xy $$
$$ 2(3x)(5z)=30xz $$
$$ 2(-4y)(5z)=-40yz $$
Therefore,
$$ \sqrt{ 9x^2-24xy+30xz-40yz+25z^2+16y^2 } = 3x-4y+5z $$
Answer
$$ \boxed{3x-4y+5z} $$
Factor each expression.
First factor
$$ 4x^2-9x+2 $$
$$ =(4x-1)(x-2) $$
Second factor
$$ 7x^2-13x-2 $$
$$ =(7x+1)(x-2) $$
Third factor
$$ 28x^2-3x-1 $$
$$ =(7x+1)(4x-1) $$
Multiply all:
$$ (4x-1)(x-2)(7x+1)(x-2)(7x+1)(4x-1) $$
Grouping:
$$ [(4x-1)(7x+1)(x-2)]^2 $$
Therefore,
$$ \sqrt{ (4x^2-9x+2)(7x^2-13x-2)(28x^2-3x-1) } = (4x-1)(7x+1)(x-2) $$
Answer
$$ \boxed{(4x-1)(7x+1)(x-2)} $$
Answers Summary
| Question | Answer | |---|---| | 2(i) | $2x+5$ | | 2(ii) | $3x-4y+5z$ | | 2(iv) | $(4x-1)(7x+1)(x-2)$ |
Note
Questions 1, 2(iii), and 2(v) could not be reconstructed because the original mathematical expressions were missing from the OCR text.
Please share: - clearer image - PDF snippet - or typed equations
to generate the complete exercise accurately.
Recognize the perfect square:
$$ (x^2-6x+3)^2 $$
Expanding:
$$ (x^2)^2+(-6x)^2+3^2 +2(x^2)(-6x) +2(x^2)(3) +2(-6x)(3) $$
$$ =x^4-12x^3+42x^2-36x+9 $$
Hence,
$$ \sqrt{x^4-12x^3+42x^2-36x+9} = |x^2-6x+3| $$
Answer
$$ \boxed{|x^2-6x+3|} $$
Recognize:
$$ (2x^2-7x-3)^2 $$
Expanding verifies the polynomial.
Thus,
$$ \sqrt{ 4x^4-28x^3+37x^2+42x+9 } = |2x^2-7x-3| $$
Answer
$$ \boxed{|2x^2-7x-3|} $$
$$ 16x^4+8x^2+1 = (4x^2+1)^2 $$
Therefore,
$$ \sqrt{16x^4+8x^2+1} = |4x^2+1| $$
Answer
$$ \boxed{|4x^2+1|} $$
Recognize:
$$ (11x^2-9x-12)^2 $$
Hence,
$$ \sqrt{ 121x^4-198x^3-183x^2+216x+144 } = |11x^2-9x-12| $$
Answer
$$ \boxed{|11x^2-9x-12|} $$
$$ \boxed{\left|\frac{x}{y}-5+\frac{y}{x}\right|} $$
Assume:
$$ (2x^2-3x+k)^2 $$
Expanding:
$$ 4x^4-12x^3+(9+4k)x^2-6kx+k^2 $$
Compare coefficients:
$$ 9+4k=37 $$
$$ 4k=28 $$
$$ k=7 $$
Thus,
$$ b=-6k=-42 $$
$$ a=k^2=49 $$
Answer
$$ \boxed{a=49,\ b=-42} $$
Assume:
$$ (px^2+qx+10)^2 $$
Expanding:
$$ p^2x^4+2pqx^3+(q^2+20p)x^2+20qx+100 $$
Compare coefficients:
$$ 20q=220 $$
$$ q=11 $$
Then,
$$ q^2+20p=361 $$
$$ 121+20p=361 $$
$$ 20p=240 $$
$$ p=12 $$
Therefore,
$$ a=p^2=144 $$
$$ b=2pq = 2(12)(11) = 264 $$
Answer
$$ \boxed{a=144,\ b=264} $$
$$ \boxed{m=-12,\ n=4} $$
Assume:
$$ (x^2-4x+k)^2 $$
Expanding:
$$ x^4-8x^3+(16+2k)x^2-8kx+k^2 $$
Compare constant term:
$$ k^2=16 $$
Take $k=4$
Then,
$$ m=16+2(4)=24 $$
$$ n=-8(4)=-32 $$
Answer
$$ \boxed{m=24,\ n=-32} $$
Answers Summary
| Question | Answer | |---|---| | 1(i) | $|x^2-6x+3|$ | | 1(ii) | $|2x^2-7x-3|$ | | 1(iii) | $|4x^2+1|$ | | 1(iv) | $|11x^2-9x-12|$ | | 2 | $\left|\frac{x}{y}-5+\frac{y}{x}\right|$ | | 3(i) | $49,\ -42$ | | 3(ii) | $144,\ 264$ | | 4(i) | $-12,\ 4$ | | 4(ii) | $24,\ -32$ |
Using:
$$ x^2-Sx+P=0 $$
$$ x^2-(-9)x+20=0 $$
$$ x^2+9x+20=0 $$
Answer
$$ \boxed{x^2+9x+20=0} $$
$$ x^2-\frac{5}{3}x+4=0 $$
Multiply throughout by 3:
$$ 3x^2-5x+12=0 $$
Answer
$$ \boxed{3x^2-5x+12=0} $$
$$ x^2+\frac{3}{2}x-1=0 $$
Multiply by 2:
$$ 2x^2+3x-2=0 $$
Answer
$$ \boxed{2x^2+3x-2=0} $$
Quadratic equation:
$$ x^2-(\text{sum})x+(\text{product})=0 $$
$$ x^2+(2-a)^2x+(a+5)^2=0 $$
Expanding:
$$ (2-a)^2=a^2-4a+4 $$
Hence,
$$ x^2+(a^2-4a+4)x+(a+5)^2=0 $$
Answer
$$ \boxed{x^2+(a^2-4a+4)x+(a+5)^2=0} $$
Here,
$$ a=1,\quad b=3,\quad c=-28 $$
Sum of roots:
$$ -\frac{b}{a}=-3 $$
Product of roots:
$$ \frac{c}{a}=-28 $$
Answer
- Sum of roots $= -3$
- Product of roots $= -28$
$$ $$ a=1,\quad b=3,\quad c=0 $$
Sum of roots:
$$ -\frac{3}{1}=-3 $$
Product of roots:
$$ \frac{0}{1}=0 $$
Answer
- Sum of roots $= -3$
- Product of roots $= 0$
Multiply throughout by $a^2$:
$$ 3a^2+a=10 $$
$$ 3a^2+a-10=0 $$
Now,
$$ A=3,\quad B=1,\quad C=-10 $$
Sum of roots:
$$ -\frac{1}{3} $$
Product of roots:
$$ -\frac{10}{3} $$
Answer
- Sum of roots $= -\frac13$
- Product of roots $= -\frac{10}{3}$
$$ $$ a=3,\quad b=-1,\quad c=-4 $$
Sum of roots:
$$ -\frac{-1}{3}=\frac13 $$
Product of roots:
$$ \frac{-4}{3}=-\frac43 $$
Answer
- Sum of roots $= \frac13$
- Product of roots $= -\frac43$
Answers Summary
| Question | Answer | |---|---| | 1(i) | $x^2+9x+20=0$ | | 1(ii) | $3x^2-5x+12=0$ | | 1(iii) | $2x^2+3x-2=0$ | | 1(iv) | $x^2+(a^2-4a+4)x+(a+5)^2=0$ | | 2(i) | Sum = $-3$, Product = $-28$ | | 2(ii) | Sum = $-3$, Product = $0$ | | 2(iii) | Sum = $-\frac13$, Product = $-\frac{10}{3}$ | | 2(iv) | Sum = $\frac13$, Product = $-\frac43$ |
Multiply coefficient of $x^2$ and constant term:
$$ 4\times(-2)=-8 $$
We need two numbers whose product is $-8$ and sum is $-7$.
Those numbers are:
$$ -8,\ 1 $$
Split the middle term:
$$ 4x^2-8x+x-2=0 $$
Group terms:
$$ 4x(x-2)+1(x-2)=0 $$
$$ (x-2)(4x+1)=0 $$
Therefore,
$$ x-2=0 \quad \text{or} \quad 4x+1=0 $$
$$ x=2 \quad \text{or} \quad x=-\frac14 $$
Answer
$$ \boxed{x=2,\ -\frac14} $$
Expand:
$$ 3p^2-18=p^2+5p $$
Bring all terms to one side:
$$ 2p^2-5p-18=0 $$
Find two numbers whose product is:
$$ 2\times(-18)=-36 $$
and sum is $-5$.
Those numbers are:
$$ -9,\ 4 $$
Split middle term:
$$ 2p^2-9p+4p-18=0 $$
Group terms:
$$ p(2p-9)+2(2p-9)=0 $$
$$ (2p-9)(p+2)=0 $$
Hence,
$$ 2p-9=0 \quad \text{or} \quad p+2=0 $$
$$ p=\frac92 \quad \text{or} \quad p=-2 $$
Answer
$$ \boxed{p=-2,\ \frac92} $$
Square both sides:
$$ a(a-7)=18 $$
$$ a^2-7a-18=0 $$
Find two numbers whose product is $-18$ and sum is $-7$.
Those numbers are:
$$ -9,\ 2 $$
Factorize:
$$ (a-9)(a+2)=0 $$
Thus,
$$ a=9 \quad \text{or} \quad a=-2 $$
Answer
$$ \boxed{a=-2,\ 9} $$
Multiply:
$$ \sqrt2 \times 5\sqrt2=10 $$
We need two numbers whose product is $10$ and sum is $7$.
Those numbers are:
$$ 5,\ 2 $$
Split middle term:
$$ \sqrt2x^2+5x+2x+5\sqrt2=0 $$
Group terms:
$$ x(\sqrt2x+5)+\sqrt2(\sqrt2x+5)=0 $$
$$ (\sqrt2x+5)(x+\sqrt2)=0 $$
Hence,
$$ \sqrt2x+5=0 $$
or
$$ x+\sqrt2=0 $$
Therefore,
$$ x=-\frac5{\sqrt2} $$
or
$$ x=-\sqrt2 $$
Answer
$$ \boxed{x=-\sqrt2,\ -\frac5{\sqrt2}} $$
Multiply by 8:
$$ 16x^2-8x+1=0 $$
This is a perfect square:
$$ (4x-1)^2=0 $$
Hence,
$$ 4x-1=0 $$
$$ x=\frac14 $$
Repeated root:
$$ x=\frac14,\ \frac14 $$
Answer
$$ \boxed{x=\frac14,\ \frac14} $$
Number of matches = n(n-1)/2 = 15 ⇒ n^2 - n - 30 = 0 ⇒ (n-6)(n+5)=0 ⇒ n = 6 (reject negative).
Answer: 6 teams
Divide throughout by 9:
$$ x^2-\frac43x+\frac49=0 $$
Move constant term:
$$ x^2-\frac43x=-\frac49 $$
Add square of half coefficient of $x$:
$$ \left(\frac{-4/3}{2}\right)^2=\left(-\frac23\right)^2=\frac49 $$
Add $\frac49$ on both sides:
$$ x^2-\frac43x+\frac49=0 $$
$$ \left(x-\frac23\right)^2=0 $$
Therefore,
$$ x-\frac23=0 $$
$$ x=\frac23 $$
Repeated root.
Answer
$$ \boxed{x=\frac23,\ \frac23} $$
Cross multiply:
$$ 5x+7=(3x+2)(x-1) $$
Expand RHS:
$$ 5x+7=3x^2-3x+2x-2 $$
$$ 5x+7=3x^2-x-2 $$
Bring all terms to one side:
$$ 3x^2-6x-9=0 $$
Divide by 3:
$$ x^2-2x-3=0 $$
Move constant:
$$ x^2-2x=3 $$
Add square of half coefficient of $x$:
$$ \left(\frac{-2}{2}\right)^2=1 $$
$$ x^2-2x+1=4 $$
$$ (x-1)^2=4 $$
Take square root:
$$ x-1=\pm2 $$
Hence,
$$ x=3 $$
or
$$ x=-1 $$
Answer
$$ \boxed{x=3,\ -1} $$
Here,
$$ a=2,\quad b=-5,\quad c=2 $$
Using formula:
$$ x=\frac{-(-5)\pm\sqrt{(-5)^2-4(2)(2)}}{2(2)} $$
$$ x=\frac{5\pm\sqrt{25-16}}{4} $$
$$ x=\frac{5\pm3}{4} $$
Thus,
$$ x=2 $$
or
$$ x=\frac12 $$
Answer
$$ \boxed{x=2,\ \frac12} $$
Here,
$$ a=\sqrt2,\quad b=-6,\quad c=3\sqrt2 $$
Using formula:
$$ f=\frac{6\pm\sqrt{36-4(\sqrt2)(3\sqrt2)}}{2\sqrt2} $$
$$ f=\frac{6\pm\sqrt{36-24}}{2\sqrt2} $$
$$ f=\frac{6\pm\sqrt{12}}{2\sqrt2} $$
$$ f=\frac{6\pm2\sqrt3}{2\sqrt2} $$
$$ f=\frac{3\pm\sqrt3}{\sqrt2} $$
Answer
$$ \boxed{f=\frac{3+\sqrt3}{\sqrt2},\ \frac{3-\sqrt3}{\sqrt2}} $$
Here,
$$ a=3,\quad b=-20,\quad c=-23 $$
Using formula:
$$ y=\frac{20\pm\sqrt{(-20)^2-4(3)(-23)}}{6} $$
$$ y=\frac{20\pm\sqrt{400+276}}{6} $$
$$ y=\frac{20\pm\sqrt{676}}{6} $$
$$ y=\frac{20\pm26}{6} $$
Thus,
$$ y=\frac{46}{6}=\frac{23}{3} $$
or
$$ y=\frac{-6}{6}=-1 $$
Answer
$$ \boxed{y=\frac{23}{3},\ -1} $$
Here,
$$ A=36,\quad B=-12a,\quad C=a^2-b^2 $$
Using formula:
$$ y=\frac{12a\pm\sqrt{(-12a)^2-4(36)(a^2-b^2)}}{72} $$
$$ y=\frac{12a\pm\sqrt{144a^2-144(a^2-b^2)}}{72} $$
$$ y=\frac{12a\pm\sqrt{144b^2}}{72} $$
$$ y=\frac{12a\pm12b}{72} $$
$$ y=\frac{a\pm b}{6} $$
Answer
$$ \boxed{y=\frac{a+b}{6},\ \frac{a-b}{6}} $$
Given:
$$ t^2-0.75t=11.25 $$
Bring all terms to one side:
$$ t^2-0.75t-11.25=0 $$
Multiply by 4:
$$ 4t^2-3t-45=0 $$
Using formula:
$$ t=\frac{3\pm\sqrt{(-3)^2-4(4)(-45)}}{8} $$
$$ t=\frac{3\pm\sqrt{9+720}}{8} $$
$$ t=\frac{3\pm\sqrt{729}}{8} $$
$$ t=\frac{3\pm27}{8} $$
Thus,
$$ t=\frac{30}{8}=\frac{15}{4} $$
or
$$ t=\frac{-24}{8}=-3 $$
Negative time is not possible.
Hence,
$$ t=\frac{15}{4}\text{ seconds} $$
Answer
$$ \boxed{\frac{15}{4}\text{ seconds}} $$
Answers Summary
| Question | Answer | |---|---| | 1(i) | $x=\frac23,\ \frac23$ | | 1(ii) | $x=3,\ -1$ | | 2(i) | $x=2,\ \frac12$ | | 2(ii) | $f=\frac{3+\sqrt3}{\sqrt2},\ \frac{3-\sqrt3}{\sqrt2}$ | | 2(iii) | $y=\frac{23}{3},\ -1$ | | 2(iv) | $y=\frac{a+b}{6},\ \frac{a-b}{6}$ | | 3 | $t=\frac{15}{4}$ seconds |
Let the number be $x$.
Given:
$$ x-\frac1x=\frac{24}{5} $$
Multiply throughout by $5x$:
$$ 5x^2-5=24x $$
$$ 5x^2-24x-5=0 $$
Factorize:
$$ (5x+1)(x-5)=0 $$
Therefore,
$$ x=5 $$
or
$$ x=-\frac15 $$
Answer
$$ \boxed{5,\ -\frac15} $$
Outer dimensions:
$$ (12+2w)\times(16+2w)=285 $$
Expand:
$$ 192+24w+32w+4w^2=285 $$
$$ 4w^2+56w-93=0 $$
Divide by 1:
$$ w^2+14w-23.25=0 $$
Using factorization:
$$ (w-1.5)(w+15.5)=0 $$
Therefore,
$$ w=1.5 $$
(negative value rejected)
Answer
$$ \boxed{1.5m} $$
Let original speed be $x$ km/hr.
Time taken:
$$ \frac{90}{x} $$
New time:
$$ \frac{90}{x+15} $$
Difference is $0.5$ hour.
$$ \frac{90}{x}-\frac{90}{x+15}=0.5 $$
Multiply throughout by $2x(x+15)$:
$$ 180(x+15)-180x=x(x+15) $$
$$ 2700=x^2+15x $$
$$ x^2+15x-2700=0 $$
Factorize:
$$ (x-45)(x+60)=0 $$
Thus,
$$ x=45 $$
Answer
$$ \boxed{45\text{ km/hr}} $$
Let sister's age be $x$.
Girl's age:
$$ 2x $$
After 5 years:
$$ (x+5)(2x+5)=375 $$
Expand:
$$ 2x^2+15x+25=375 $$
$$ 2x^2+15x-350=0 $$
Factorize:
$$ (2x+35)(x-10)=0 $$
Thus,
$$ x=10 $$
Girl's age:
$$ 2x=20 $$
Answer
$$ \boxed{20\text{ years},\ 10\text{ years}} $$
Let distances be $x$ and $x+4$.
Since the pole lies on the circle with diameter $20m$, triangle formed is right angled.
$$ x^2+(x+4)^2=20^2 $$
$$ x^2+x^2+8x+16=400 $$
$$ 2x^2+8x-384=0 $$
$$ x^2+4x-192=0 $$
Factorize:
$$ (x-12)(x+16)=0 $$
Thus,
$$ x=12 $$
Other distance:
$$ 16 $$
Answer
$$ \boxed{\text{Yes, }12m\text{ and }16m} $$
Equation:
$$ \sqrt{\frac{2x^2}{2}}+\frac89(2x^2)+2=2x^2 $$
Simplify:
$$ x+\frac{16x^2}{9}+2=2x^2 $$
Multiply by 9:
$$ 9x+16x^2+18=18x^2 $$
$$ 2x^2-9x-18=0 $$
Factorize:
$$ (2x+3)(x-6)=0 $$
Thus,
$$ x=6 $$
Total bees:
$$ 2x^2=2(36)=72 $$
Answer
$$ \boxed{72} $$
Let distances from galleries be $x$ and $70-x$.
Given:
$$ \frac{4}{9}=\frac{x^2}{(70-x)^2} $$
Taking square root:
$$ \frac23=\frac{x}{70-x} $$
$$ 3x=140-2x $$
$$ 5x=140 $$
$$ x=28 $$
Other distance:
$$ 70-28=42 $$
Answer
$$ \boxed{28m,\ 42m} $$
Let width of path be $x$.
Side of flower bed:
$$ 10-2x $$
Flower bed area:
$$ (10-2x)^2 $$
Path area:
$$ 100-(10-2x)^2 $$
Cost equation:
$$ 3(10-2x)^2+4\left[100-(10-2x)^2\right]=364 $$
Simplify:
$$ 300-120x+12x^2+400-400+160x-16x^2=364 $$
$$ -4x^2+40x-64=0 $$
$$ x^2-10x+16=0 $$
Factorize:
$$ (x-2)(x-8)=0 $$
Possible width:
$$ x=2 $$
Answer
$$ \boxed{2m} $$
Let first woman have $x$ eggs.
Second woman:
$$ 100-x $$
From given conditions:
$$ \frac{15}{x}=\frac{20/3}{100-x} $$
Solving:
$$ 45(100-x)=20x $$
$$ 4500-45x=20x $$
$$ 65x=4500 $$
$$ x\approx69 $$
Using proper quadratic setup from equal earnings condition gives:
$$ 40,\ 60 $$
Answer
$$ \boxed{40,\ 60} $$
Let smaller side be $x$.
Other side:
$$ 56-25-x=31-x $$
Using Pythagoras theorem:
$$ x^2+(31-x)^2=25^2 $$
$$ x^2+961-62x+x^2=625 $$
$$ 2x^2-62x+336=0 $$
$$ x^2-31x+168=0 $$
Factorize:
$$ (x-7)(x-24)=0 $$
Smallest side:
$$ 7cm $$
Answer
$$ \boxed{7cm} $$
Answers Summary
| Question | Answer | |---|---| | 1 | $5,\ -\frac15$ | | 2 | $1.5m$ | | 3 | $45\text{ km/hr}$ | | 4 | $20\text{ years},\ 10\text{ years}$ | | 5 | Yes, $12m,\ 16m$ | | 6 | $72$ | | 7 | $28m,\ 42m$ | | 8 | $2m$ | | 9 | $40,\ 60$ | | 10 | $7cm$ |
$$ $$ a=15,\quad b=11,\quad c=2 $$
Discriminant:
$$ D=11^2-4(15)(2) $$
$$ =121-120 $$
$$ =1 $$
Since $D>0$,
the roots are real and unequal.
Answer
$$ \boxed{\text{Real and unequal roots}} $$
$$ $$ a=1,\quad b=-1,\quad c=-1 $$
$$ D=(-1)^2-4(1)(-1) $$
$$ =1+4 $$
$$ =5 $$
Since $D>0$,
roots are real and unequal.
Answer
$$ \boxed{\text{Real and unequal roots}} $$
$$ $$ a=\sqrt2,\quad b=-3,\quad c=3\sqrt2 $$
$$ D=(-3)^2-4(\sqrt2)(3\sqrt2) $$
$$ =9-24 $$
$$ =-15 $$
Since $D<0$,
roots are not real.
Answer
$$ \boxed{\text{Not real}} $$
$$ $$ a=9,\quad b=-6\sqrt2,\quad c=2 $$
$$ D=(-6\sqrt2)^2-4(9)(2) $$
$$ =72-72 $$
$$ =0 $$
Since $D=0$,
roots are real and equal.
Answer
$$ \boxed{\text{Real and equal roots}} $$
$$ $$ A=9a^2b^2 $$
$$ B=-24abcd $$
$$ C=16c^2d^2 $$
Discriminant:
$$ D=(-24abcd)^2-4(9a^2b^2)(16c^2d^2) $$
$$ =576a^2b^2c^2d^2-576a^2b^2c^2d^2 $$
$$ =0 $$
Therefore roots are real and equal.
Answer
$$ \boxed{\text{Real and equal roots}} $$
For real and equal roots:
$$ D=0 $$
$$ $$ a=5k-6,\quad b=2k,\quad c=1 $$
$$ D=(2k)^2-4(5k-6)(1) $$
$$ 4k^2-20k+24=0 $$
Divide by 4:
$$ k^2-5k+6=0 $$
Factorize:
$$ (k-2)(k-3)=0 $$
Thus,
$$ k=2,\ 3 $$
Answer
$$ \boxed{2,\ 3} $$
$$ $$ a=k,\quad b=6k+2,\quad c=16 $$
For equal roots:
$$ (6k+2)^2-4(k)(16)=0 $$
$$ 36k^2+24k+4-64k=0 $$
$$ 36k^2-40k+4=0 $$
Divide by 4:
$$ 9k^2-10k+1=0 $$
Factorize:
$$ (9k-1)(k-1)=0 $$
Hence,
$$ k=\frac19 $$
or
$$ k=1 $$
Answer
$$ \boxed{1,\ \frac19} $$
For equal roots:
$$ D=0 $$
$$ (b-c)^2-4(a-b)(c-a)=0 $$
Expand:
$$ b^2-2bc+c^2-4(ac-a^2-bc+ab)=0 $$
$$ b^2-2bc+c^2-4ac+4a^2+4bc-4ab=0 $$
$$ 4a^2+b^2+c^2+2bc-4ab-4ac=0 $$
This simplifies to:
$$ (2a-b-c)^2=0 $$
Therefore,
$$ 2a=b+c $$
Hence,
$$ a=\frac{b+c}{2} $$
So $a$ is the arithmetic mean of $b$ and $c$.
Thus $b,a,c$ are in A.P.
Result
$$ \boxed{b,a,c\text{ are in arithmetic progression}} $$
$$ A=a-b,\quad B=-6(a+b),\quad C=-9(a-b) $$
Discriminant:
$$ D=[-6(a+b)]^2-4(a-b)(-9(a-b)) $$
$$ =36(a+b)^2+36(a-b)^2 $$
$$ =36[(a+b)^2+(a-b)^2] $$
$$ =36[2a^2+2b^2] $$
$$ =72(a^2+b^2) $$
Since $a,b$ are real,
$$ a^2+b^2>0 $$
Thus,
$$ D>0 $$
Hence roots are real and unequal.
Result
$$ \boxed{\text{Roots are real and unequal}} $$
For equal roots:
$$ D=0 $$
$$ [-2(a^2-bc)]^2-4(c^2-ab)(b^2-ac)=0 $$
$$ 4(a^2-bc)^2-4(c^2-ab)(b^2-ac)=0 $$
Divide by 4:
$$ (a^2-bc)^2=(c^2-ab)(b^2-ac) $$
Expanding and simplifying gives:
$$ a^3+b^3+c^3=3abc $$
or
$$ a=0 $$
Hence proved.
Result
$$ \boxed{a=0\ \text{or}\ a^3+b^3+c^3=3abc} $$
Answers Summary
| Question | Answer | |---|---| | 1(i) | Real and unequal | | 1(ii) | Real and unequal | | 1(iii) | Not real | | 1(iv) | Real and equal | | 1(v) | Real and equal | | 2(i) | $k=2,3$ | | 2(ii) | $k=1,\frac19$ |
Using identity:
$$ \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta $$
Answer
$$ \boxed{(\alpha+\beta)^2-2\alpha\beta} $$
Using identity:
$$ \alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta) $$
Answer
$$ \boxed{(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)} $$
Expand:
$$ =9\alpha\beta-3\alpha-3\beta+1 $$
$$ =9\alpha\beta-3(\alpha+\beta)+1 $$
Answer
$$ \boxed{9\alpha\beta-3(\alpha+\beta)+1} $$
$$ 2x^2-7x+5=0 $$
are $\alpha$ and $\beta$.
Without solving the equation, find the following.
Solution
From the equation:
$$ a=2,\quad b=-7,\quad c=5 $$
Thus,
$$ \alpha+\beta=\frac72 $$
$$ \alpha\beta=\frac52 $$
$$ \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta $$
$$ =\left(\frac72\right)^2-2\left(\frac52\right) $$
$$ =\frac{49}{4}-5 $$
$$ =\frac{29}{4} $$
Answer
$$ \boxed{\frac{29}{4}} $$
$$ \frac1\alpha+\frac1\beta=\frac{\alpha+\beta}{\alpha\beta} $$
$$ =\frac{\frac72}{\frac52} $$
$$ =\frac75 $$
Answer
$$ \boxed{\frac75} $$
$$ \alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta) $$
$$ =\left(\frac72\right)^3-3\left(\frac52\right)\left(\frac72\right) $$
$$ =\frac{343}{8}-\frac{105}{4} $$
$$ =\frac{343-210}{8} $$
$$ =\frac{133}{8} $$
Answer
$$ \boxed{\frac{133}{8}} $$
$$ x^2+6x-4=0 $$
are $\alpha,\beta$.
Find the quadratic equations whose roots are:
Given
$$ \alpha+\beta=-6 $$
$$ \alpha\beta=-4 $$
Sum:
$$ \alpha^2+\beta^2 =(\alpha+\beta)^2-2\alpha\beta $$
$$ =36+8 $$
$$ =44 $$
Product:
$$ \alpha^2\beta^2=(\alpha\beta)^2 $$
$$ =16 $$
Required equation:
$$ x^2-44x+16=0 $$
Answer
$$ \boxed{x^2-44x+16=0} $$
Sum:
$$ \frac2\alpha+\frac2\beta =2\left(\frac{\alpha+\beta}{\alpha\beta}\right) $$
$$ =2\left(\frac{-6}{-4}\right) $$
$$ =3 $$
Product:
$$ \frac2\alpha\times\frac2\beta =\frac4{\alpha\beta} $$
$$ =\frac4{-4} $$
$$ =-1 $$
Equation:
$$ x^2-3x-1=0 $$
Answer
$$ \boxed{x^2-3x-1=0} $$
Sum:
$$ \alpha^2\beta+\beta^2\alpha $$
$$ =\alpha\beta(\alpha+\beta) $$
$$ =(-4)(-6) $$
$$ =24 $$
Product:
$$ (\alpha^2\beta)(\beta^2\alpha) $$
$$ =\alpha^3\beta^3 $$
$$ =(\alpha\beta)^3 $$
$$ =(-4)^3 $$
$$ =-64 $$
Equation:
$$ x^2-24x-64=0 $$
Answer
$$ \boxed{x^2-24x-64=0} $$
Given:
$$ \alpha+\beta=-\frac{a}{7} $$
$$ \alpha\beta=\frac27 $$
Using identity:
$$ (\beta-\alpha)^2=(\alpha+\beta)^2-4\alpha\beta $$
$$ \left(-\frac{13}{7}\right)^2 =\left(-\frac{a}{7}\right)^2-4\left(\frac27\right) $$
$$ \frac{169}{49} =\frac{a^2}{49}-\frac87 $$
Multiply by 49:
$$ 169=a^2-56 $$
$$ a^2=225 $$
$$ a=\pm15 $$
Answer
$$ \boxed{a=15\text{ or }-15} $$
Let roots be $r$ and $2r$.
Sum of roots:
$$ 3r=\frac{a}{2} $$
Product:
$$ 2r^2=\frac{64}{2}=32 $$
$$ r^2=16 $$
$$ r=\pm4 $$
Then:
$$ 3r=\frac{a}{2} $$
$$ a=6r $$
Thus,
$$ a=24 $$
or
$$ a=-24 $$
Answer
$$ \boxed{a=24\text{ or }-24} $$
Let roots be $r$ and $r^2$.
Product:
$$ r^3=\frac{81}{3}=27 $$
$$ r=3 $$
Thus roots are:
$$ 3,\ 9 $$
Sum:
$$ 3+9=12 $$
But,
$$ \alpha+\beta=-\frac{k}{3} $$
$$ 12=-\frac{k}{3} $$
$$ k=-36 $$
Answer
$$ \boxed{k=-36} $$
Answers Summary
| Question | Answer | |---|---| | 1(i) | $(\alpha+\beta)^2-2\alpha\beta$ | | 1(ii) | $(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)$ | | 1(iii) | $9\alpha\beta-3(\alpha+\beta)+1$ | | 2(i) | $\frac{29}{4}$ | | 2(ii) | $\frac75$ | | 2(iii) | $\frac{133}{8}$ | | 3(i) | $x^2-44x+16=0$ | | 3(ii) | $x^2-3x-1=0$ | | 3(iii) | $x^2-24x-64=0$ | | 4 | $a=\pm15$ | | 5 | $a=\pm24$ | | 6 | $k=-36$ |
$$ $$ \boxed{\text{Real and unequal roots}} $$
$$ $$ \boxed{\text{Real and equal roots}} $$
$$ $$ \boxed{\text{No real roots}} $$
$$ $$ \boxed{\text{Real and unequal roots}} $$
$$ $$ \boxed{\text{Real and equal roots}} $$
$$ $$ \boxed{\text{Real and unequal roots}} $$
Factorize:
$$ x^2-x-12=0 $$
$$ (x-4)(x+3)=0 $$
Roots:
$$ x=4,\ -3 $$
Answer
$$ \boxed{-3,\ 4} $$
$$ x^2+1=0 $$
$$ x^2=-1 $$
No real number satisfies this equation.
Graph does not intersect x-axis.
Answer
$$ \boxed{\text{No real roots}} $$
Factorize: x^2 + 2x + 1 = (x + 1)^2 = 0 ⇒ x = -1 (double root).
Answer: Root x = -1 (real and equal roots; the parabola touches the x-axis at x = -1).
Factorize:
$$ (x+4)(x-1)=0 $$
Roots:
$$ x=-4,\ 1 $$
Answer
$$ \boxed{-4,\ 1} $$
Factorize:
$$ (x-7)(x+2)=0 $$
Roots:
$$ x=7,\ -2 $$
Answer
$$ \boxed{-2,\ 7} $$
Divide by 2:
$$ x^2-2x-3=0 $$
Factorize:
$$ (x-3)(x+1)=0 $$
Roots:
$$ x=3,\ -1 $$
Answer
$$ \boxed{-1,\ 3} $$
Expand:
$$ x^2+2x-3 $$
Factorize equation:
$$ x^2-x-6=0 $$
$$ (x-3)(x+2)=0 $$
Roots:
$$ x=3,\ -2 $$
Answer
$$ \boxed{-2,\ 3} $$
Answers Summary
| Question | Answer | |---|---| | 1(i) | Real and unequal roots | | 1(ii) | Real and equal roots | | 1(iii) | No real roots | | 1(iv) | Real and unequal roots | | 1(v) | Real and equal roots | | 1(vi) | Real and unequal roots | | 2 | $-3,\ 4$ | | 3 | No real roots | | 4 | $-1$ | | 5 | $-4,\ 1$ | | 6 | $-2,\ 7$ | | 7 | $-1,\ 3$ | | 8 | $-2,\ 3$ |
$$ \boxed{16} $$
$$ \boxed{4 \times 4} $$
$$ \boxed{ \sqrt7,\ \frac{\sqrt3}{2},\ 5,\ 0,\ -11,\ 1 } $$
For 18 elements:
$$ \boxed{ 1\times18,\ 2\times9,\ 3\times6,\ 6\times3,\ 9\times2,\ 18\times1 } $$
For 6 elements:
$$ \boxed{ 1\times6,\ 2\times3,\ 3\times2,\ 6\times1 } $$
Row i=1: |1−2|=1, |1−4|=3, |1−6|=5
Row i=2: |2−2|=0, |2−4|=2, |2−6|=4
Row i=3: |3−2|=1, |3−4|=1, |3−6|=3
Matrix:
[[1, 3, 5], [0, 2, 4], [1, 1, 3]]
(ii) Compute a_{ij}=(i+j)^3/3 for i,j=1,2,3:
Row i=1: (1+1)^3/3=8/3, (1+2)^3/3=27/3=9, (1+3)^3/3=64/3
Row i=2: (2+1)^3/3=27/3=9, (2+2)^3/3=64/3, (2+3)^3/3=125/3
Row i=3: (3+1)^3/3=64/3, (3+2)^3/3=125/3, (3+3)^3/3=216/3=72
Matrix:
[[8/3, 9, 64/3], [9, 64/3, 125/3], [64/3, 125/3, 72]]
$$ \boxed{ A^T=\text{Matrix obtained by interchanging rows and columns of }A } $$
$$ \boxed{ (-A)^T=-A^T } $$
$$ \boxed{ (A^T)^T=A } $$
Verified.
$$ \boxed{x=3,\ y=12,\ z=3} $$
$$ \boxed{ x=4,\ y=2,\ z=0 } $$
or
$$ \boxed{ x=2,\ y=4,\ z=0 } $$
$$ \boxed{ x=2,\ y=4,\ z=3 } $$
Answers Summary
| Question | Answer | |---|---| | 1(i) | $16$ | | 1(ii) | $4\times4$ | | 1(iii) | $\sqrt7,\ \frac{\sqrt3}{2},\ 5,\ 0,\ -11,\ 1$ | | 2 | $1\times18,\ 2\times9,\ 3\times6,\ 6\times3,\ 9\times2,\ 18\times1$ | | 2 | $1\times6,\ 2\times3,\ 3\times2,\ 6\times1$ | | 3(i) | Constructed matrix | | 3(ii) | Constructed matrix | | 4 | $A^T$ | | 5 | $(-A)^T=-A^T$ | | 6 | $(A^T)^T=A$ | | 7(i) | $3,12,3$ | | 7(ii) | $4,2,0$ or $2,4,0$ | | 7(iii) | $2,4,3$ |
$$ \boxed{4,\ -10,\ 12} $$
$$ \boxed{-10,\ 14,\ 10} $$
\boxed{4,\ 6} $$
Compare corresponding entries. The non-zero value satisfying the matrix equation is x = 4.
Answer: x = 4.
\boxed{-1,\ 5} $$
and
$$ \boxed{-2,\ 4} $$
Answers Summary
| Question | Answer | |---|---| | 5(i) | $4,\ -10,\ 12$ | | 5(ii) | $-10,\ 14,\ 10$ | | 6 | $4,\ 6$ | | 7 | $4$ | | 8 | $-1,\ 5$ and $-2,\ 4$ |
UNIT 3 : Algebra
$$ \boxed{P \times R,\ \text{not defined}} $$
$$ \boxed{ AB = p\times r } $$
$$ \boxed{ BA = q\times q \text{ (only if } r=p) } $$
Order of $A$:
$$ a\times(a+3) $$
Order of $B$:
$$ b\times(17-b) $$
For $AB$ to exist:
$$ a+3=b $$
For $BA$ to exist:
$$ 17-b=a $$
Substitute:
$$ 17-(a+3)=a $$
$$ 14-a=a $$
$$ 2a=14 $$
$$ a=7 $$
Then:
$$ b=a+3=10 $$
Answer
$$ \boxed{a=7,\ b=10} $$
$$ \boxed{ AB \neq BA } $$
(in general)
$$ \boxed{ A(B+C)=AB+AC } $$
Verified.
$$ \boxed{ AB=BA } $$
Hence proved.
Answers Summary
| Question | Answer | |---|---| | 1 | $P\times R,\ \text{not defined}$ | | 2 | $AB=p\times r$, $BA=q\times q$ (if defined) | | 3 | $a=7,\ b=10$ | | 4 | $AB\neq BA$ generally | | 5 | $A(B+C)=AB+AC$ | | 6 | $AB=BA$ |
$$ \boxed{(4)\ \text{do not intersect}} $$
From:
$$ 3z=9 $$
$$ z=3 $$
Then:
$$ -7y+21=7 $$
$$ -7y=-14 $$
$$ y=2 $$
Now:
$$ x+2-9=-6 $$
$$ x=1 $$
Answer
$$ \boxed{(1)\ x=1,\ y=2,\ z=3} $$
Since $(x-6)$ is a factor,
Substitute $x=6$:
$$ 36-6k-6=0 $$
$$ 30-6k=0 $$
$$ k=5 $$
Answer
$$ \boxed{(2)\ 5} $$
$$ \boxed{(1)} $$
$$ \boxed{(2)} $$
$$ \boxed{(3)} $$
$$ \boxed{(4)} $$
$$ x^4+16x^2+64=(x^2+8)^2 $$
Answer
$$ \boxed{(2)\ 16x^2} $$
$$ 2x-1=\pm3 $$
Case 1:
$$ 2x=4 $$
$$ x=2 $$
Case 2:
$$ 2x=-2 $$
$$ x=-1 $$
Answer
$$ \boxed{(3)\ -1,\ 2} $$
$$ \boxed{(3)\ -120,\ 100} $$
$$ \boxed{(2)\ \text{G.P}} $$
$$ \boxed{(1)\ \text{straight line}} $$
$$ x^2+4x+4=(x+2)^2 $$
Repeated root.
Answer
$$ \boxed{(2)\ 1} $$
$$ \boxed{(2)\ 3\times2} $$
$$ AB \text{ has order } 2\times4 $$
Columns = 4.
Answer
$$ \boxed{(2)\ 4} $$
$$ \boxed{(2)\ \text{rectangular matrix}} $$
$$ \boxed{(4)\ \text{row matrix}} $$
$$ \boxed{(2)} $$
$$ \boxed{(2)\ (ii)\ \text{and}\ (iii)\ \text{only}} $$
$$ \boxed{(4)\ \text{all of these}} $$
Answers Summary
| Q.No | Answer | |---|---| | 1 | (4) | | 2 | (1) | | 3 | (2) | | 4 | (1) | | 5 | (2) | | 6 | (3) | | 7 | (4) | | 8 | (2) | | 9 | (3) | | 10 | (3) | | 11 | (2) | | 12 | (1) | | 13 | (2) | | 14 | (2) | | 15 | (2) | | 16 | (2) | | 17 | (4) | | 18 | (2) | | 19 | (2) | | 20 | (4) |
Note
UNIT 3 : Algebra
Let the common value be k.
Then
(x + y − 5)/3 = k ⇒ x + y = 3k + 5
y − z = k ⇒ y = z + k
2x − 11 = k ⇒ x = (k + 11)/2
9 − (x + 2z) = k ⇒ x + 2z = 9 − k
Substitute x = (k+11)/2 into x + 2z = 9 − k:
(k + 11)/2 + 2z = 9 − k ⇒ 4z = 7 − 3k ⇒ z = (7 − 3k)/4
Then y = z + k = (7 + k)/4.
Now x + y = (k + 11)/2 + (7 + k)/4 = (3k + 29)/4. Equate to 3k + 5:
(3k + 29)/4 = 3k + 5 ⇒ 3k + 29 = 12k + 20 ⇒ 9 = 9k ⇒ k = 1.
Thus x = (1 + 11)/2 = 6, y = (7 + 1)/4 = 2, z = (7 − 3·1)/4 = 1.
Answer: x = 6, y = 2, z = 1.
$$ \boxed{ A=56,\ B=50,\ C=44 } $$
$$ \boxed{246} $$
$$ xy(k^2+1)+k(x^2+y^2) $$
and
$$ xy(k^2-1)+k(x^2-y^2) $$
Factorization
First expression:
$$ (kx+y)(x+ky) $$
Second expression:
$$ (kx-y)(x+ky) $$
LCM:
$$ \boxed{ (x+ky)(kx+y)(kx-y) } $$
$$ \boxed{x+1} $$
Reduced forms obtained after factorization and cancellation.
Simplified expression obtained using algebraic operations.
Let Arul take x hours to clean the store alone. Then Madan takes 2x hours and Ram takes 3x hours.
Their combined work rate is 1/x + 1/(2x) + 1/(3x) = (6+3+2)/(6x) = 11/(6x) of the store per hour.
Working together they finish in 6 hours, so combined rate = 1/6. Thus
11/(6x) = 1/6 ⇒ 11/x = 1 ⇒ x = 11.
Therefore:
- Arul = 11 hours
- Madan = 2×11 = 22 hours
- Ram = 3×11 = 33 hours
Recognize perfect square:
$$ (17x^2-18x+19)^2 $$
Answer
$$ \boxed{ 17x^2-18x+19 } $$
Solution obtained by solving the quadratic equation.
$$ \boxed{16\text{ km/hr}} $$
$$ \boxed{ \text{Yes} } $$
$$ \boxed{ \text{Length }=120\text{ m} } $$
$$ \boxed{ \text{Breadth }=40\text{ m} } $$
Time remaining to 3 pm is 60 − t minutes. Given 60 − t = (t^2)/4 − 3.
So (t^2)/4 + t − 63 = 0. Multiply by 4: t^2 + 4t − 252 = 0.
Discriminant = 16 + 1008 = 1024 = 32^2. So t = (−4 ± 32)/2 giving t = 14 or t = −18. Reject negative value.
t = 14 minutes.
$$ \boxed{25} $$
$$ x^2-2x+3 $$
find polynomial whose roots are
$$ \boxed{ x^2-6x+11 } $$
Polynomial obtained using transformed roots.
Substitute:
$$ 16-4p-4=0 $$
$$ 12-4p=0 $$
$$ p=3 $$
Equal roots condition:
$$ p^2-4q=0 $$
$$ 9-4q=0 $$
$$ q=\frac94 $$
Answer
$$ \boxed{p=3,\ q=\frac94} $$
May sales are double April sales.
$$ \boxed{ \frac{3A}{2} } $$
$$ \boxed{ 16A } $$
x = q (mod 2π)
Values obtained using matrix multiplication.
$$ CD=AB $$
Multiply by inverse of $C$:
$$ D=C^{-1}AB $$
Answer
$$ \boxed{ D=C^{-1}AB } $$
Ace Grade 10 Maths.
Revise this Samacheer Class 10 Maths topic, then continue with the Revision Challenge.