Question

find the roots of 4x²+3x+5=0 by the method of completing the square​

Answer 1

Solution :

The given equation is

$4x^{2}+3x+5=0$

$\to \small{(2x)^{2}+2.2x.\frac{3}{4}+(\frac{3}{4})^{2}+5-(\frac{3}{4})^{2}=0}$

$\to (2x+\frac{3}{4})^{2}+5-\frac{9}{16}=0$

$\to (2x+\frac{3}{4})^{2}+\frac{80-9}{16}=0$

$\to (2x+\frac{3}{4})^{2}+\frac{71}{9}=0$

$\to \small{(2x+\frac{3}{4})^{2}-(\frac{\sqrt{71}}{3}i)^{2}=0,\:\:i^{2}=-1}$

$\to \small{(2x+\frac{3}{4}+\frac{\sqrt{71}}{3}i)(2x+\frac{3}{4}-\frac{\sqrt{71}}{3}i)=0}$

Either $2x+\frac{3}{4}+\frac{\sqrt{71}}{3}i=0$

$\to 2x=-\frac{3}{4}-\frac{\sqrt{71}}{3}i$

$\to x=-\frac{9+4\sqrt{71}i}{24}$

Or, $2x+\frac{3}{4}-\frac{\sqrt{71}}{3}i=0$

$\to 2x=-\frac{3}{4}+\frac{\sqrt{71}}{3}i$

$\to x=-\frac{9-4\sqrt{71}i}{24}$

Thus, the required solution be

    $\boxed{x=-\frac{9\pm 4\sqrt{71}i}{24}}$

Note : The given equation has no real roots because the value of the discriminant is less than 0.

For the standard quadratic equation $ax^{2}+bx+c=0,\:a\neq 0$

$D=b^{2}-4ac$

$=3^{2}-4.4.5=9-80=-71<0$

{ From the given equation ↑ }