Let the cube side be a (>0). New cuboid sides: a+1,a+2,a+3. Increase in volume: (a+1)(a+2)(a+3)-a^3 = 52. Expand: a^3+6a^2+11a+6 - a^3 = 6a^2+11a+6 = 52. So 6a^2+11a-46=0. Discriminant Δ = 11^2+4·6·46 = 121+1104=1225=(35)^2. Thus a = (-11±35)/(12) gives a=2 or a=-23/6. Reject negative. So a=2 and cuboid volume = (3)(4)(5)=60.
60
(i) (x-1)(x-2)(x-3)=x^3-6x^2+11x-6. (ii) (x-1)^2(x+2)= (x^2-2x+1)(x+2)=x^3-3x+2. (iii) (x-2)(x-1/2)(x-1)= x^3-(7/2)x^2+(7/2)x-1; clearing denominators: 2x^3-7x^2+7x-2=0.
(i) x^3-6x^2+11x-6=0 (ii) x^3-3x+2=0 (iii) 2x^3-7x^2+7x-2=0
(i) 2α, 2β, 2γ,
(ii) $\frac{1}{α}$, $\frac{1}{β}$, $\frac{1}{γ}$
(iii) – α, – β, – γv
(i) Given that α, β, γ are the roots of x 3 + 2x 2 + 3x + 4 = 0
Compare with x 3 + bx 2 + cx + d = 0
b = 2, c = 3, d = 4
α + β + γ = -6 = -2
αβ + βγ + γα = c = 3
αβγ = -d = -4
Given roots are 2α, 2β, 2γ
2α + 2β + 2γ = 2 (α + β + γ)
= 2 (-2)
= -4
(2α) (2β) + (2β) (2γ) + (2γ) (2α) = (4αβ + 4βγ + 4γα)
= 4(αβ + βγ + γα)
= 4(3)
= 12
(2α) (2β) (2γ) = 8(αβγ)
= 8(-4)
= -32
The equation is
x 3 – x 2 (2α + 2β + 2γ) + x (4αβ + 4βγ + 4γα) – 8 (αβγ) = 0
⇒ x 3 – x 2 (-4) + x (12) – (-32) = 0
⇒ x 3 + 4x 2 + 12x + 32 = 0
(ii) The new roots are $\frac{1}{α}$, $\frac{1}{β}$, $\frac{1}{γ}$
⇒ 4x³ + 3x² + 2x + 1 = 0
(iii) The given roots are -α, -β, -γ
The cubic equation is
x 3 – x 2 (-α – β – γ) + x (αβ + βγ + γα) + (αβγ) = 0
⇒ x 3 + x 2 (α + β + γ) + x (αβ + βγ + γα) + (αβγ) = 0
⇒ x 3 + x 2 (-2) + x (3) – 4 = 0
⇒ x 3 – 2x 2 + 3x – 4 = 0
⇒ x 3 + x 2 (α + β + γ) + x (αβ + βγ + γα) + (αβγ) = 0 ⇒ x 3 + x 2 (-2) + x (3) – 4 = 0 ⇒ x 3 – 2x 2 + 3x – 4 = 0
For 3x^3-16x^2+23x-6=0, product of all roots = -d/a = -(-6)/3 = 2. If two roots have product 1, the third root = 2/1 =2, so x=2 is a root. Divide polynomial by (x-2): quotient 3x^2-10x+3. Solve 3x^2-10x+3=0 ⇒ x=(10±√(100-36))/6=(10±8)/6 → x=3 or x=1/3. So roots 2,3,1/3 (two of which multiply to 1: 3·1/3=1).
Roots: 3, 2, 1/3
The given equation is 2x 4 – 8x 3 + 6x 2 – 3 = 0.
(÷ 2) ⇒ x 4 – 4x 3 + 3x 2 – $\frac{3}{2}$ = 0
Let the roots be α, β, γ, δ
α + β + γ + δ = -b = 4
(αβ + βγ + γδ + αδ + αγ + βδ) = c = 3
αβγ + βγδ + γδα = -d = 0
αβγδ = $\frac{-3}{2}$
To Find α 2 + β 2 + γ 2 + δ 2 = (α + β + γ + δ) 2 – 2 (αβ + βγ + γδ + αδ + αγ + βδ)
= (4) 2 – 2(3)
= 16 – 6
= 10
= (4) 2 – 2(3) = 16 – 6 = 10
Let the roots are 3α, 2α, β
sum of the roots are
3α + 2α + β = 9
5α + β = 9 ………. (1)
Product of two roots
3α(2α) + 2α(β) + β(3α) = 14
6α² + 5αβ = 14 ……… (2)
Product of three roots
(3α) (2α)β = -24
α²β = -4 ………. (3)
(1) ⇒ β = 9 – 5 α
(2) ⇒ 6α² + 5α (9 – 5α) = 14
6α² + 45α – 25α² = 14
-19α² + 45α – 14 = 0
19α² – 45α + 14 = 0
(α – 2) (α – $\frac{7}{19}$) = 0
α = 2 or α = $\frac{7}{19}$
If α = 2, β = 9 – 5 (α) = 9 – 5(2) = 9 – 10 = -1
roots are 3α, 2α, β
3(2), 2(2), -1 (i,e.,) 6, 4, -1
If α = $\frac{7}{19}$, β = 9 – 5($\frac{7}{19}$) = ($\frac{136}{19}$)
roots are 3α, 2α, β (i,e.,) $\frac{21}{19}$, $\frac{14}{19}$, $\frac{136}{19}$
3(2), 2(2), -1 (i,e.,) 6, 4, -1 If α = $\frac{7}{19}$, β = 9 – 5($\frac{7}{19}$) = ($\frac{136}{19}$) roots are 3α, 2α, β (i,e.,) $\frac{21}{19}$, $\frac{14}{19}$, $\frac{136}{19}$
Σ α/(βγ) = α/(βγ)+β/(γα)+γ/(αβ) = (α^2+β^2+γ^2)/(αβγ). Now α^2+β^2+γ^2 = (α+β+γ)^2 -2(αβ+βγ+γα) = (-b/a)^2 -2(c/a) = (b^2 -2ac)/a^2. Also αβγ = -d/a. Hence Σ α/(βγ) = [(b^2-2ac)/a^2] / (-d/a) = (b^2-2ac)/(-ad) = (2ac-b^2)/(ad).
Σ α/(βγ) = (2ac - b^2)/(ad)
pq = ($\frac{-5}{2}$)(4) = -10
x² – (p + q)x + pq = 0
x² – $\frac{3}{2}$x – 10 = 0
2x² – 3x – 20 = 0
x² – (p + q)x + pq = 0 x² – $\frac{3}{2}$x – 10 = 0 2x² – 3x – 20 = 0
Let the height of the tree = 12
length of the cut part = x³
Length of left out part = $\sqrt[3]{x^{3}}$
= x
Given x + x³ = 12
x³ + x – 12 = 0
Which is required mathematical problem
Given x + x³ = 12 x³ + x – 12 = 0 Which is required mathematical problem
For ax^2+bx+c with a=2,b=k,c=k, discriminant Δ = k^2 - 4·2·k = k^2 -8k = k(k-8). So (i) Δ>0 ⇒ k<0 or k>8: two distinct real roots. (ii) Δ=0 ⇒ k=0 or k=8: equal real roots. (iii) 0<k<8 ⇒ Δ<0: two complex conjugate roots.
Discriminant Δ = k^2 - 8k. Roots real and distinct if Δ>0 ⇔ k(k-8)>0 ⇒ k<0 or k>8. Real and equal if k=0 or k=8. Complex conjugate if 0<k<8.
Complex conjugate 2-3i is also a root. Minimal polynomial is (x-(2+3i))(x-(2-3i)) = (x-2)^2 + 9 = x^2 -4x +13.
x^2 - 4x + 13 = 0
Same as previous: include the conjugate 2-3i; minimal polynomial (x-(2+3i))(x-(2-3i)) = x^2 -4x +13.
x^2 - 4x + 13 = 0
Let the root be √5 – √3,
Another root is √5 + √3
Sum of the roots = √5 – √3 + √5 + √3 = 2√5
Product of roots = (√5 – √3) (√5 + √3)
√5² – √3² = 5 – 3 = 2
x² – (SR)x + PR = 0
x² – 2√5 x + 2 = 0 which is not rational co-efficient.
to make rational co-efficient
(x² + 2√5 x + 2) (x² + 2 + 2√5 x) = 0
(x² + 2)² – (2√5x)² = 0
x 4 + 4 + 4x² – 20x² = 0
⇒ x 4 – 16x² + 4 = 0 is a rational co-efficient polynomial equation.
(x² + 2)² – (2√5x)² = 0 x 4 + 4 + 4x² – 20x² = 0 ⇒ x 4 – 16x² + 4 = 0 is a rational co-efficient polynomial equation.
Let parabola y = ax^2 + bx + c and line y = px + q. Intersection points satisfy ax^2 + bx + c = px + q ⇒ ax^2 + (b-p)x + (c-q) = 0, a quadratic in x. A quadratic has at most two distinct real roots, so the line and parabola intersect in at most two points.
Intersection reduces to solving a quadratic; at most two real solutions.
The given equation is 2x 3 – x 2 – 18x + 9 = 0
$x^{3}-\frac{x^{2}}{2}-9 x+\frac{9}{2}=0$
Let the roots be α, -α, β
α – α + β = $-\left(\frac{-1}{2}\right)$
$\Rightarrow \beta=\frac{1}{2}$
(α) (-α) (β) = $\frac{-9}{2}$
$\Rightarrow-\alpha^{2}\left(\frac{1}{2}\right)=\frac{-9}{2}$
α 2 = 9
α = ±3
The roots are 3, -3, $\frac { 1 }{ 2 }$
α 2 = 9 α = ±3 The roots are 3, -3, $\frac { 1 }{ 2 }$
Given the roots are in AP
Let the roots be a – d, a, a + d
Given the roots are in AP Let the roots be a – d, a, a + d
3 + 3λ + 3λ² = 13λ
3λ² + 3λ – 13λ + 3 = 0
3λ² – 10λ + 3 = 0
(λ – 3) (3λ – 1) = 0
λ = 6 or λ = $\frac{1}{3}$
3λ² – 10λ + 3 = 0 (λ – 3) (3λ – 1) = 0 λ = 6 or λ = $\frac{1}{3}$
Given cubic equation
2x³ – 6x² + 3x + k = 0
Let the roots be α, β, γ
Given α = 2(β + γ)
β + γ = $\frac{α}{2}$ ………. (1)
Sum of roots α(β + γ) = 3
From (1) α + $\frac{α}{2}$ = 3
$\frac{3α}{2}$ = 3 ⇒ α = 2
Again αβ + βγ + γα = $\frac{3}{2}$
α = 2 ⇒ 2β + βγ + 2γ = $\frac{3}{2}$
from (1) 2($\frac{α}{2}$) + βγ = $\frac{3}{2}$
βγ = $\frac{3}{2}$ – 2 = $\frac{3-4}{2}$ = $\frac{-1}{2}$
βγ = $\frac{-1}{2}$ ………… (2)
product of roots α β γ = –$\frac{k}{2}$
α = 2 ⇒ 2βγ = –$\frac{k}{2}$
βγ = –$\frac{k}{4}$ ………… (3)
product of roots α β γ = –$\frac{k}{2}$ α = 2 ⇒ 2βγ = –$\frac{k}{2}$ βγ = –$\frac{k}{4}$ ………… (3)
(i) Given that 1 + 2i, √3
Another roots be 1 – 2i, -√3
sum of roots = 1 + 2i + 1 – 2i
product roots = (1 + 2i)(1 – 2i)
1² + 2² = 1 + 4 = 5
x² – 2x + 5 = 0
(ii) sum of roots = √3 – √3
product roots = (√3)(-√3)
x² – 0x – 3 = 0
x² – 3 = 0
(x² – 2x + 5)(x² – 3) = x 4 – 2x³ + 2x² + 6x – 15
x 6 – 3x 5 – 5x 4 + 22x³ – 39x² – 39x + 135
= (x 4 – 2x³ + 2x² + 6x – 15) (x² + px – 9)
Equate of co-efficient of x on both sides
-39 = -54 – 15 p
-39 + 54 = -15 p
15 = -15 p
p = -1
∴ x² – x – 9 = 0
15 = -15 p p = -1 ∴ x² – x – 9 = 0
(i) 2x³ – 9x² + 10x = 3,
2x³ – 9x² + 10x – 3 = 0v
(i) 2x³ – 9x² + 10x = 3
2x³ – 9x² + 10x – 3 = 0
sum of the coefficients 2 – 9 + 10 – 3 = 0
∴ x = 1 is one of the roots.
2x² – 7x + 3 = 0
(x – 3)(2x – 1) = 0
x = 3 or x = $\frac{1}{2}$
roots are 3, $\frac{1}{2}$, 1
(ii) 8x³ – 2x² – 7x + 3 = 0
sum of the alternative coefficients are equal
8 – 7 = -2 + 3
1 = 1
∴ (x + 1) is a factor.
8x² – 10x + 3 = 0
(4x – 3) (2x – 1) = 0
4x = 3 (or) 2x = 1
x = $\frac{3}{4}$ (or) $\frac{1}{2}$
∴ The roots are $\frac{3}{4}$, $\frac{1}{2}$, -1
4x = 3 (or) 2x = 1 x = $\frac{3}{4}$ (or) $\frac{1}{2}$ ∴ The roots are $\frac{3}{4}$, $\frac{1}{2}$, -1
x 4 – 14x² + 45 = 0v
Put t = x² ⇒ (x²)² – 14x² + 45 = 0
t² – 14t + 45 = 0
t² – 9t – 5t + 45 = 0
t(t – 9) – 5(t – 9) = 0
(t – 9)(t – 5) = 0
t – 9 = 0 or t – 5 = 0
t = 9 or t = 5
x² = 9 or x² = 5
x = ± 3 or y = ±√5
The roots are ±3, ±√5
x² = 9 or x² = 5 x = ± 3 or y = ±√5 The roots are ±3, ±√5
(i) (x – 5) (x – 7) (x + 6)(x + 4) = 504v
(x – 5) (x + 4) (x – 7) (x + 6) = 504
(x 2 – x – 20) (x 2 – x – 42) = 504
Let y = (x 2 – x)
(y – 20) (y – 42) = 504
⇒ y 2 – 42y – 20y + 840 = 504
⇒ y 2 – 62y + 336 = 0
⇒ (y – 56) (y – 6) = 0
⇒ (y – 56) = 0 or (y – 6) = 0
⇒ x 2 – x – 56 = 0 or x 2 – x – 6 = 0
⇒ (x – 8) (x + 7) = 0 or (x – 3) (x + 2) = 0
⇒ x = 8, -7 or x = 3, -2
The roots are 8, -7, 3, -2
(ii) (x – 4)(x – 2)(x- 7)(x + 1) = 16
(x – 4) (x – 7) (x – 2) (x + 1) = 16
⇒ (x – 4) (x – 2) (x – 7) (x + 1) = 16
⇒ (x 2 – 6x + 8) (x 2 – 6x – 7) = 16
Let x 2 – 6x = y
(y + 8)(y – 7) = 16
⇒ y 2 – 7y + 8y – 56 – 16 = 0
⇒ y 2 + y – 72 = 0
⇒ (y + 9) (y – 8) = 0
y + 9 = 0
x 2 – 6x + 9 = 0
(x – 3) 2 = 0
x = 3, 3
or
y – 8 = 0
x 2 – 6x – 8 = 0
or y – 8 = 0 x 2 – 6x – 8 = 0
(2x – 1)(x + 3)(x – 2)(2x + 3) + 20 = 0v
(x + 3)(x – 2) (2x – 1)(2x + 3) + 20 = 0
(x² + x – 6) (4x² + 6x – 2x – 3) + 20 = 0
(x² + x – 6) (4x² + 4x – 3) + 20 = 0
(x² + x – 6) (x² + 4x – $\frac{3}{4}$) + 20 = 0
(÷4) (x² + x – 6) (x² + x – $\frac{3}{4}$) + $\frac{20}{4}$ = 0
Put y = x² + x
4y² – 27y – 38 = 0
4y² – 8y – 19y + 38 = 0
(4y – 19) (y – 2) = 0
y = 2 (or) y = $\frac{19}{4}$
x² + x = 2 (or) x² + x = $\frac{19}{4}$
x² + x – 2 = 0 (or) 4x² + 4x = 19
(x + 2)(x – 1) = 0 (or) 4x² + 4x – 19 = 0
x = -2 or x = 1
x² + x – 2 = 0 (or) 4x² + 4x = 19 (x + 2)(x – 1) = 0 (or) 4x² + 4x – 19 = 0 x = -2 or x = 1
(i) sin² x – 5 sin x + 4 = 0v
sin 2 x – 5sinx + 4 = 0
Let y = sin x
(y 2 – 5y + 4 = 0
(y – 1) (y – 4) = 0
(y – 1) = o or (y – 4) = 0
y = 1 or y = 4
sin x = 1 or sin x = 4 [not possible since sin x ≤ 1]
sin x = sin $\frac{\pi}{2}$
x = nπ + (-1) n α, n ∈ z.
x = nπ + (-1) n $\frac{\pi}{2}$
(ii) 12x³ + 8x = 29x² – 4 = 0
12x³ – 29x² + 8x + 4 = 0
12x² – 5x – 2 = 0
12x² – 8x + 3x – 2 = 0
4x(3x – 2) + 1(3x – 2) = 0
(3x – 2)(4x + 1) = 0
3x = 2, 4x = -1
x = $\frac{2}{3}$ or x = –$\frac{1}{4}$
The roots are 2, $\frac{2}{3}$, –$\frac{1}{4}$.
3x = 2, 4x = -1 x = $\frac{2}{3}$ or x = –$\frac{1}{4}$ The roots are 2, $\frac{2}{3}$, –$\frac{1}{4}$.
(i) 2x³ – x² – 1 = 0v
Sum of co-efficients = 2 – 1 – 1 = 0
⇒ x = 1 is a factor.
Which is imaginary root
∴ x = 1 is rational root.
(ii) x 8 – 3x + 1 = 0
a n = 1; a 0 = 1
If $\frac{p}{q}$ is a root of the polynomial. (p, q) = 1
By rational root theorem, it has no rational roots.
a n = 1; a 0 = 1 If $\frac{p}{q}$ is a root of the polynomial. (p, q) = 1 By rational root theorem, it has no rational roots.
Put k = $\frac{3}{2n}$ ∴ 8x k – 8x -k = 63
8x k – $\frac{8}{x^k}$ = 63
8x 2k – 8 = 63x k
8x 2k – 63x k – 8 = 0
(x k – 8) (8x k + 1) = 0
x k – 8 = 0
x k = 8
x $\frac{3}{2n}$ = 8
x = 8 $\frac{3}{2n}$ (2³) $\frac{3}{2n}$ = (2²) n = 4 n
∴ x = 4 n is a root of the equation.
x $\frac{3}{2n}$ = 8 x = 8 $\frac{3}{2n}$ (2³) $\frac{3}{2n}$ = (2²) n = 4 n ∴ x = 4 n is a root of the equation.
put $\sqrt{\frac{x}{a}}$ = y
put $\sqrt{\frac{x}{a}}$ = y
(i) 6x 4 – 35x³ + 62x² – 35x + 6 = 0v
This is Type I even degree reciprocal equation. Hence it can be rewritten as
(1) ⇒ 6(y² – 2) – 35y + 62 = 0
6y² – 12 – 35y + 62 = 0
6y² – 35y + 50 = 0
2x² + 2 – 5x = 0 (or) 3x² + 3 = 10x
2x² – 5x + 2 = 0 (or) 3x² – 10x + 3 = 0
x = 2, $\frac{1}{2}$ (or) x = 3, $\frac{1}{3}$
Roots are 2, $\frac{1}{2}$, 3 and $\frac{1}{3}$
(ii) x 4 + 3x³ – 3x – 1 = 0
(x – 1) and (x + 1) is a factor.
(x – 1)(x + 1)(x² + 3x + 1) = 0
x – 1 = 0 (or) x + 1 = 0 (or) x² + 3x + 1 = 0
x = 1 (or) x = -1 (or) x² + 3x = -1
(x – 1)(x + 1)(x² + 3x + 1) = 0 x – 1 = 0 (or) x + 1 = 0 (or) x² + 3x + 1 = 0 x = 1 (or) x = -1 (or) x² + 3x = -1
4 x – 3 (2 x+2 ) + 2 5 = 0
⇒ (2 2 ) x – 3(2 x. 2 2 ) + 2 5 = 0
(2 2 ) x – 12. 2 x + 32 = 0
Let y = 2 x
y 2 – 12y + 32 = 0
⇒ (y – 4) (y – 8) = 0
y – 4 = 0 or y – 8 = 0
Case (i): 2 x = 4
⇒ 2 x = (2) 2
⇒ x = 2
Case (ii): 2x = 8
⇒ 2 x = (2) 3
⇒ x = 3
∴ The roots are 2, 3
⇒ 2 x = (2) 3 ⇒ x = 3 ∴ The roots are 2, 3
6x 4 – 5x³ – 38x² – 5x + 6 = 0
6(y² – 2) – 5y – 38 = 0
6y² – 12 – 5y – 38 = 0
6y² – 5y – 50 = 0
6y² – 20y + 15y – 50 = 0
2y(3y – 10) + 5(3y – 10) = 0
(3y – 10)(2y + 5) = 0
3y = 10, 2y = -5
3x² + 3 = 10x, 2x² + 2 = -5x
3x² – 10x + 3 = 0, 2x² + 5x + 2 = 0
(x – 3)(3x – 1) = 0, (2x + 1)(x + 1) = 0
x = 3, x = $\frac{1}{3}$ or x = $\frac{-1}{2}$, x= -2
The roots are 3, $\frac{1}{3}$, -2, $\frac{-1}{2}$.
(x – 3)(3x – 1) = 0, (2x + 1)(x + 1) = 0 x = 3, x = $\frac{1}{3}$ or x = $\frac{-1}{2}$, x= -2 The roots are 3, $\frac{1}{3}$, -2, $\frac{-1}{2}$.
P(x) = 9x 9 – 4x 8 + 4x 7 – 3x 6 + 2x 5 + x 3 + 7x 2 + 7x + 2
The number of sign changes in P(x) is 4.
∴P(x) has at most 4 positive roots.
P(-x) = -9x 9 – 4x 8 – 4x 7 – 3x 6 – 2x 5 – x 3 + 7x 2 – 7x + 2
The number of sign changes in P(-x) is 3.
P(x) has almost 3 negative roots.
Since the difference between the number of sign changes in co-efficient P(-x) and the number of negative roots of the polynomial P(x) is even.
The number of negative roots = at most 2.
P(x) has almost 3 negative roots. Since the difference between the number of sign changes in co-efficient P(-x) and the number of negative roots of the polynomial P(x) is even. The number of negative roots = at most 2.
y = x² – 5x + 6
x = 1, y = 1 – 5 + 6 = 2
x = 2, y = 4 – 10 + 6 = 0
x = 0, y = 6
x = 3, y = 9 – 15 + 6 = 0
x = -1, y = 1 + 5 + 6 = 12
x = 4, y = 16 – 20 + 6 = 2
(1, 2), (0, 6), (-1, 12)
P(x) = (x² + 5x + 6) (x² – 5x + 16)
= x 4 – 5x³ + 16x² – 5x³ + 25x² – 80x + 6x² – 30x + 96 = 0
= x 4 – 10x³ + 47x² – 110x + 96 = 0
It has two sign changes
∴ It has two positive real roots
P(-x) = x 4 + 10x³ + 47x² + 110x + 96
It has no sign change, no negative real roots
y = x 2 – 5x + 16
P(-x) = x 4 + 10x³ + 47x² + 110x + 96 It has no sign change, no negative real roots y = x 2 – 5x + 16
P(x) = x 9 – 5x 5 + 4x 4 + 2x 2 + 1
(i) The number of sign changes in P(x) is 2. The number of positive roots is atmost 2.
(ii) P(-x) = -x 9 + 5x 5 + 4x 4 + 2x 2 + 1. The number of sign changes in P(-x) is 1. The number of negative roots of P (x) is atmost 1. Since the difference of number of sign changes in P(-x) and number of negative zeros is even.
P(x) has one negative root.
(iii) 0 is not the zero of the polynomial P(x). So the number of real roots is almost 3.
∴ The number of imaginary roots at least 6.
P(x) has one negative root. (iii) 0 is not the zero of the polynomial P(x). So the number of real roots is almost 3. ∴ The number of imaginary roots at least 6.
P(x) = x 9 – 5x 8 – 14x 7. It has only one sign change.
∴ It has one positive roots.
P(-x) = -x 9 – 5x 8 + 14x 7. It has only one sign change.
It has one negative root.
∴ It has one positive and one negative roots.
P(-x) = -x 9 – 5x 8 + 14x 7. It has only one sign change. It has one negative root. ∴ It has one positive and one negative roots.
P(x) = x 9 + 9x 7 + 7x 5 + 5x 3 + 3x.
There is no change in the sign of P(x) and P(-x), P(x) has no positive and no negative real roots, but 0 is the root of the polynomial equation P(x).
P(x) = x 9 + 9x 7 + 7x 5 + 5x 3 + 3x. There is no change in the sign of P(x) and P(-x), P(x) has no positive and no negative real roots, but 0 is the root of the polynomial equation P(x).
x^3 + 64 = 0 ⇒ x^3 = -64 ⇒ real cube root x = -4. Hence correct option: (4) -4.
deg(f(g(x))) = deg(f)·deg(g) = m·n. So option (1) mn is correct.
By the Fundamental Theorem of Algebra a degree n polynomial (over C) has exactly n complex roots counted with multiplicity. So option (3) is correct.
For monic cubic: α+β+γ = -p, αβ+βγ+γα = q, αβγ = -r. Σ(1/α) = (αβ+βγ+γα)/(αβγ) = q/(-r) = −q/r. So option (1) is correct.
(A) -1 (B) $\frac{5}{4}$ (C) $\frac{4}{5}$ (D) 5v
(c) $\frac{4}{5}$
Hint:
By rational root of theorem,
$\frac{p}{q}$ is a root of a polynomial a 0 = -5, a n = 4 and (4, -5) = 1, then p must divide 5 and q must divide 4.
Possible values of p are +1, -1, +5, -5
Possible values of q are +1, -1, 4, -4
∴ $\frac{4}{5}$ is not possible rational roots.
$\frac{4}{5}$