Question

Find the roots of any one of the following quadratic equation
1) $8x^2-14x+5=0$ (By factorizations method)
2) $2x^2-7x+3=0$ (By completing the square methods)

Answer 1

We have,

Part (1).

$8x^2-14x+5=0$

By factorization method,

$8x^2-14x+5=0$

$8x^2-(10+4)x+5=0$

$8x^2-10x-4x+5=0$

$2x(4x-5)-1(4x-5)=0$

$(4x-5)(2x-1)=0$

If $4x-5=0$ then, $x=\frac{5}{4}$

If $2x-1=0$ then, $x=\frac{1}{2}$

Hence, this is the answer.

Part $\left(2\right)$,

$2x^2-7x+3=0$

By completing square method,

$2x^2-7x+3=0$

Divide both side by $2$ and we get,

$x^2-\frac{7}{2}x+\frac{3}{2}=0$

Adding ${\left(\frac{7}{4}\right)}^2$ both side and we get,

$x^2-\frac{7}{2}x+\frac{3}{2}+{\left(\frac{7}{4}\right)}^2=0+{\left(\frac{7}{4}\right)}^2$

$x^2-\frac{7}{2}x+{\left(\frac{7}{4}\right)}^2+\frac{3}{2}={\left(\frac{7}{4}\right)}^2$

${(x-\frac{7}{4})}^2=\frac{49}{16}-\frac{3}{2}$

${(x-\frac{7}{4})}^2=\frac{49-24}{16}$

${(x-\frac{7}{4})}^2=\frac{25}{16}$

$(x-\frac{7}{4})=\sqrt{\frac{25}{16}}$

$(x-\frac{7}{4})=\pm \frac{5}{4}$

If $+ve$ sign,

$x-\frac{7}{4}=\frac{5}{4}$

$x=\frac{7}{4}+\frac{5}{4}$

$x=\frac{7+5}{4}=\frac{12}{4}$

$x=3$

If $-vesign$

$x-\frac{7}{4}=-\frac{5}{4}$

$x=\frac{7}{4}-\frac{5}{4}$

$x=\frac{7-5}{4}$

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

Hence, this is the answer.