NCERT Solutions: Class 10 Maths Chapter 4 - Quadratic Equations

(i) Expanding gives $x^2 + 2x + 1 = 2x - 6$, i.e., $x^2 + 7 = 0$, so it is quadratic.
(ii) $x^2 - 2x = -6 + 2x$ gives $x^2 - 4x + 6 = 0$, so it is quadratic.
(iii) Expanding gives $x^2 - x - 2 = x^2 + 2x - 3$, i.e., $-3x + 1 = 0$, which is linear.
(iv) Expanding gives $2x^2 - 5x - 3 = x^2 + 5x$, i.e., $x^2 - 10x - 3 = 0$, so it is quadratic.
(v) Expanding gives $2x^2 - 7x + 3 = x^2 + 4x - 5$, i.e., $x^2 - 11x + 8 = 0$, so it is quadratic.
(vi) Expanding gives $x^2 + 3x + 1 = x^2 - 4x + 4$, i.e., $7x - 3 = 0$, which is linear.
(vii) Expanding gives $x^3 + 6x^2 + 12x + 8 = 2x^3 - 2x$, i.e., $x^3 - 6x^2 - 14x - 8 = 0$, which is cubic.
(viii) Expanding gives $x^3 - 4x^2 - x + 1 = x^3 - 6x^2 + 12x - 8$, i.e., $2x^2 - 13x + 9 = 0$, so it is quadratic.

(i) Let breadth $= x$ m. Then length $= 2x + 1$ m, and $x(2x + 1) = 528$, so $2x^2 + x - 528 = 0$.
(ii) Let the integers be $x$ and $x + 1$. Then $x(x + 1) = 306$, so $x^2 + x - 306 = 0$.
(iii) Let Rohan’s present age be $x$ years. His mother’s present age is $x + 26$. Three years from now, their ages will be $x + 3$ and $x + 29$, so $(x + 3)(x + 29) = 360$, i.e., $x^2 + 32x - 273 = 0$.
(iv) Let the speed be $x$ km/h. The original time is $\dfrac{480}{x}$ hours and the slower time is $\dfrac{480}{x - 8}$ hours. Therefore $\dfrac{480}{x - 8} = \dfrac{480}{x} + 3$, which simplifies to $x^2 - 8x - 1280 = 0$.

(i) $x^2 - 3x - 10 = (x - 5)(x + 2)$, so $x = 5, -2$.
(ii) $2x^2 + x - 6 = (2x - 3)(x + 2)$, so $x = \dfrac{3}{2}, -2$.
(iii) $\sqrt{2}x^2 + 7x + 5\sqrt{2} = (x + \sqrt{2})(\sqrt{2}x + 5)$, so $x = -\sqrt{2}, -\dfrac{5}{\sqrt{2}}$.
(iv) $2x^2 - x + \dfrac{1}{8} = 0$. Multiplying by 8 gives $16x^2 - 8x + 1 = (4x - 1)^2 = 0$, so $x = \dfrac{1}{4}$ twice.
(v) $100x^2 - 20x + 1 = (10x - 1)^2 = 0$, so $x = \dfrac{1}{10}$ twice.

Example 1(i) gives the equation $x^2 - 45x + 324 = 0$, where $x$ is the number of marbles John had. Factorising, $x^2 - 45x + 324 = (x - 36)(x - 9) = 0$, so $x = 36$ or $x = 9$. Since the total number of marbles is 45, the other person had 9 or 36 marbles.
Example 1(ii) gives the equation $x^2 - 55x + 750 = 0$, where $x$ is the number of toys produced. Factorising, $x^2 - 55x + 750 = (x - 25)(x - 30) = 0$, so $x = 25$ or $x = 30$. The cost of each toy is $55 - x$, so it is ₹30 or ₹25 respectively.

Let one number be $x$. The other is $27 - x$. Their product is 182, so $x(27 - x) = 182$, i.e., $x^2 - 27x + 182 = 0$. Factorising, $(x - 13)(x - 14) = 0$, so the numbers are 13 and 14.

Let the consecutive positive integers be $x$ and $x + 1$. Then $x^2 + (x + 1)^2 = 365$, so $2x^2 + 2x - 364 = 0$, i.e., $x^2 + x - 182 = 0$. Factorising, $(x - 13)(x + 14) = 0$. Since the integer is positive, $x = 13$, so the integers are 13 and 14.

Let the altitude be $x$ cm. Then the base is $x + 7$ cm. By Pythagoras theorem, $x^2 + (x + 7)^2 = 13^2$. This gives $2x^2 + 14x - 120 = 0$, or $x^2 + 7x - 60 = 0$. Factorising, $(x - 5)(x + 12) = 0$. Since a length cannot be negative, $x = 5$. Hence altitude $= 5$ cm and base $= 12$ cm.

Let the number of articles be $x$. Then the cost of each article is $2x + 3$ rupees. Total cost $= x(2x + 3) = 90$, so $2x^2 + 3x - 90 = 0$. Factorising, $(2x + 15)(x - 6) = 0$. Since $x$ is positive, $x = 6$. Cost of each article $= 2(6) + 3 = ₹15$.

Use the discriminant $D = b^2 - 4ac$.
(i) For $2x^2 - 3x + 5 = 0$, $D = (-3)^2 - 4(2)(5) = 9 - 40 = -31 \lt 0$, so there are no real roots.
(ii) For $3x^2 - 4\sqrt{3}x + 4 = 0$, $D = (-4\sqrt{3})^2 - 4(3)(4) = 48 - 48 = 0$, so the roots are equal. $x = \dfrac{-b}{2a} = \dfrac{4\sqrt{3}}{6} = \dfrac{2\sqrt{3}}{3}$.
(iii) For $2x^2 - 6x + 3 = 0$, $D = (-6)^2 - 4(2)(3) = 36 - 24 = 12 \gt 0$, so there are two distinct real roots. $x = \dfrac{6 \pm \sqrt{12}}{4} = \dfrac{6 \pm 2\sqrt{3}}{4} = \dfrac{3 \pm \sqrt{3}}{2}$.

For equal roots, discriminant $D = 0$.
(i) $D = k^2 - 4(2)(3) = k^2 - 24$. So $k^2 - 24 = 0$, giving $k = \pm 2\sqrt{6}$.
(ii) $kx(x - 2) + 6 = 0$ gives $kx^2 - 2kx + 6 = 0$. Here $a = k$, $b = -2k$, $c = 6$. $D = (-2k)^2 - 4(k)(6) = 4k^2 - 24k = 4k(k - 6)$. For a quadratic equation, $k \neq 0$, so $k = 6$.

Let breadth be $x$ m. Then length $= 2x$ m. Area $= 2x^2 = 800$, so $x^2 = 400$ and $x = 20$ since breadth is positive. Hence length $= 40$ m.

Let one present age be $x$ years. Then the other is $20 - x$ years. Four years ago, their ages were $x - 4$ and $16 - x$. Given $(x - 4)(16 - x) = 48$. Expanding gives $x^2 - 20x + 112 = 0$. Its discriminant is $D = (-20)^2 - 4(1)(112) = 400 - 448 = -48 \lt 0$. Since there are no real roots, the situation is not possible.

Let length be $l$ m and breadth be $b$ m. Perimeter $= 80$ gives $2(l + b) = 80$, so $l + b = 40$. Also $lb = 400$. Put $b = 40 - l$. Then $l(40 - l) = 400$, so $l^2 - 40l + 400 = 0$, i.e., $(l - 20)^2 = 0$. Hence $l = 20$ and $b = 20$.