Find the roots of the following quadratic equations by factorisation:
(i) $x^2 -3x – 10 = 0$
(ii) $2x^2 + x – 6 = 0$
(iii) $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$
(iv) $2x^2 – x + \frac{1}{8} = 0$
(v) $100x^2 – 20x+ 1 = 0$

AcademicMathematicsNCERTClass 10

To do:

We have to find the roots of the given quadratic equations by factorisation.

Solution:

(i) $x^2-3x-10=0$

$x^2-5x+2x-10=0$

$x(x-5)+2(x-5)=0$

$(x-5)(x+2)=0$

$x-5=0$ or $x+2=0$

$x=5$ or $x=-2$

Hence, the roots of the given quadratic equation are $-2$ and $5$.

(ii) $2x^2+x-6=0$

$2x^2+4x-3x-6=0$

$2x(x+2)-3(x+2)=0$

$(x+2)(2x-3)=0$

$x+2=0$ or $2x-3=0$

$x=-2$ or $2x=3$

$x=-2$ or $x=\frac{3}{2}$

Hence, the roots of the given quadratic equation are $-2$ and $\frac{3}{2}$.

(iii) $\sqrt{2}x^2+7x+5\sqrt2=0$

To factorise $\sqrt{2}x^2+7x+5\sqrt2=0$, we have to find two numbers $m$ and $n$ such that $m+n=7$ and $mn=\sqrt{2}\times(5\sqrt{2})=5(\sqrt2)^2=10$.

If $m=5$ and $n=2$, $m+n=5+2=7$ and $mn=(5)2=10$.

$\sqrt{2}x^2+5x+2x+5\sqrt2=0$

$\sqrt{2}x(x+\sqrt2)+5(x+\sqrt2)=0$

$(\sqrt{2}x+5)(x+\sqrt2)=0$

$\sqrt{2}x+5=0$ or $x+\sqrt2=0$

$\sqrt{2}x=-5$ or $x=-\sqrt2$

$x=-\frac{5}{\sqrt2}$ or $x=-\sqrt2$

The values of $x$ are $-\frac{5}{\sqrt2}$ and $-\sqrt2$.

(iv) $2x^2 – x + \frac{1}{8} = 0$

$\frac{8(2x^2)-8(x)+1}{8}=0$

$16x^2-8x+1=0(8)$

$16x^2-8x+1=0$

$16x^2-4x-4x+1=0$

$4x(4x-1)-1(4x-1)=0$

$(4x-1)(4x-1)=0$

$4x-1=0$ or $4x-1=0$

$4x=1$ or $4x=1$

$x=\frac{1}{4}$ or $x=\frac{1}{4}$

Hence, the roots of the given quadratic equation are $\frac{1}{4}$ and $\frac{1}{4}$.

(v) $100x^2 – 20x+ 1 = 0$

$100x^2-10x-10x+1=0$

$10x(10x-1)-1(10x-1)=0$

$(10x-1)(10x-1)=0$

$10x-1=0$ or $10x-1=0$

$10x=1$ or $10x=1$

$x=\frac{1}{10}$ or $x=\frac{1}{10}$

Hence, the roots of the given quadratic equation are $\frac{1}{10}$ and $\frac{1}{10}$.

raja
Updated on 10-Oct-2022 13:20:12

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