Find the roots of the following quadratic equations by the factorisation method:
$ 2 x^{2}+\frac{5}{3} x-2=0 $

To do:

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

Solution:

(i) \( 2 x^{2}+\frac{5}{3} x-2=0 \)

Multiplying by 3 on both sides, we get,

$6 x^{2}+5 x-6 =0$

$6x^{2}+(9 x-4 x)-6 =0$

$6 x^{2}+9 x-4 x-6=0$

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

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

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

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

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

(ii) \( \frac{2}{5} x^{2}-x-\frac{3}{5}=0 \)

Multiplying by 5 on both sides, we get,

$2x^2-5x-3=0$

$2 x^{2}-(6 x-x)-3=0$

$2 x^{2}-6 x+x-3 =0$

$2 x(x-3)+1(x-3) =0$

$(x-3)(2 x+1)=0$

$x-3=0$ or $2x+1=0$

$x=3$ or $x=-\frac{1}{2}$

Hence, the roots of the given quadratic equation are $-\frac{1}{2}, 3$. 

(iii) \( 3 \sqrt{2} x^{2}-5 x-\sqrt{2}=0 \)

$3 \sqrt{2} x^{2}-(6 x-x)-\sqrt{2}=0$

$3 \sqrt{2} x^{2}-6 x+x-\sqrt{2}=0$

$3 \sqrt{2} x^{2}-3 \sqrt{2}(\sqrt{2})x+x-\sqrt{2}=0$

$3 \sqrt{2} x(x-\sqrt{2})+1(x-\sqrt{2})=0$

$(x-\sqrt{2})(3 \sqrt{2} x+1)=0$

$x-\sqrt{2}=0$ or $3 \sqrt{2} x+1=0$

$x =\sqrt{2}$ or $x=-\frac{1}{3 \sqrt{2}}$

$x =\sqrt{2}$ or $x=-\frac{\sqrt{2}}{3\sqrt{2}\times \sqrt2}$

$x =\sqrt{2}$ or $x=-\frac{\sqrt{2}}{6}$

Hence, the roots of the given quadratic equation are $\sqrt{2}, -\frac{\sqrt{2}}{6}$. 

(iv) \( 3 x^{2}+5 \sqrt{5} x-10=0 \)

$3 x^{2}+6 \sqrt{5} x-\sqrt{5} x-2 \sqrt{5}(\sqrt{5})=0$

$3 x(x+2 \sqrt{5})-\sqrt{5}(x+2 \sqrt{5})=0$

$(x+2 \sqrt{5})(3 x-\sqrt{5})=0$

$x+2 \sqrt{5}=0$ or $3 x-\sqrt{5} =0$

$x=-2 \sqrt{5}$ or $x=\frac{\sqrt{5}}{3}$

Hence, the roots of the given quadratic equation are $-2 \sqrt{5}, \frac{\sqrt{5}}{3}$.  

(v) \( 21 x^{2}-2 x+\frac{1}{21}=0 \)

Multiplying by 21 on both sides, we get,

$441 x^{2}-42 x+1=0$

$441 x^{2}-(21 x+21 x)+1=0$

$441 x^{2}-21 x-21 x+1=0$

$21 x(21 x-1)-1(21 x-1)=0$

$(21 x-1)(21 x-1)=0$

$21 x-1=0$ or $21 x-1=0$

$x=\frac{1}{21}$ or $x=\frac{1}{21}$

Hence, the roots of the given quadratic equation are $\frac{1}{21}, \frac{1}{21}$.   

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Updated on: 10-Oct-2022

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