Find the roots of the following quadratic

Question:

Find the roots of the following quadratic equations by the factorisation method.

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

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

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

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

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

Solution:

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

On multiplying by 3 on both sides, we get

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

$\Rightarrow \quad 6 x^{2}+(9 x-4 x)-6=0 \quad$ [by splitting the middle term]

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

$\Rightarrow \quad 3 x(2 x+3)-2(2 x+3)=0$

$\Rightarrow \quad(2 x+3)(3 x-2)=0$

Now, $2 x+3=0$

$\Rightarrow \quad x=-\frac{3}{2}$

and $3 x-2=0$

$\Rightarrow$  $x=\frac{2}{3}$

Hence, the roots of the equation $6 x^{2}+5 x-6=0$ are $\frac{-3}{2}$ and $\frac{2}{3}$.

(ii) Given equation is $\frac{2}{5} x^{2}-x-\frac{3}{5}=0$.

On multiplying by 5 on both sides, we get

$\Rightarrow$ $2 x^{2}-5 x-3=0$

$2 x^{2}-(6 x-x)-3=0$ [by splitting the middle term]

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

$\Rightarrow \quad 2 x(x-3)+1(x-3)=0$

$\Rightarrow \quad(x-3)(2 x+1)=0$

Now, $x-3=0 \Rightarrow x=3$

and $2 x+1=0$

$\Rightarrow$ $x=-\frac{1}{2}$

Hence, the roots of the equation $2 x^{2}-5 x-3=0$ are $-\frac{1}{2}$ and 3 .

(iii) Given equation is $3 \sqrt{2} x^{2}-5 x-\sqrt{2}=0$.

$3 \sqrt{2} x^{2}-(6 x-x)-\sqrt{2}=0$ [by splitting the middle term]

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

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

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

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

Now, $x-\sqrt{2}=0 \Rightarrow x=\sqrt{2}$

and $\quad 3 \sqrt{2} x+1=0$

$\Rightarrow$ $x=-\frac{1}{3 \sqrt{2}}=\frac{-\sqrt{2}}{6}$

Hence, the roots of the equation $3 \sqrt{2} x^{2}-5 x-\sqrt{2}=0$ are $-\frac{\sqrt{2}}{6}$ and $\sqrt{2}$.

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

$3 x^{2}+6 \sqrt{5} x-\sqrt{5} x-2 \sqrt{5} \cdot \sqrt{5}=0$ [by splitting the middle term]

$\Rightarrow \quad 3 x^{2}+6 \sqrt{5} x-\sqrt{5} x-10=0$

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

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

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

Now, $x+2 \sqrt{5}=0$

$\Rightarrow \quad x=-2 \sqrt{5}$ and $3 x-\sqrt{5}=0$

$\Rightarrow \quad x=\frac{\sqrt{5}}{3}$

Hence, the roots of the equation $3 x^{2}+5 \sqrt{5} x-10=0$ are $-2 \sqrt{5}$ and $\frac{\sqrt{5}}{3}$.

(v) Given equation is $21 x^{2}-2 x+\frac{1}{21}=0$.

On multiplying by 21 on both sides, we get

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

$441 x^{2}-(21 x+21 x)+1=0$ [by splitting the middle term]

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

$\Rightarrow \quad 21 x(21 x-1)-1(21 x-1)=0$

$\Rightarrow \quad(21 x-1)(21 x-1)=0$

Now, $21 x-1=0 \Rightarrow x=\frac{1}{21}$ and $21 x-1=0$

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

Hence, the roots of the equation $441 x^{2}-42 x+1=0$ are $\frac{1}{21}$ and $\frac{1}{21}$.