Question

Use the quadratic formula to solve for the roots in the following equation.

Answer 1

Answer:

The roots of $4 x^{2}+5 x+2=2 x^{2}+7 x-1$  are $\frac{2+2 i \sqrt{5}}{2}$ and $\frac{2-2 i \sqrt{5}}{2}$

Solution:

The equation given is,

$4 x^{2}+5 x+2=2 x^{2}+7 x-1$

Simplifying we get,

$4 x^{2}+5 x+2=2 x^{2}+7 x-1$

$4x^2- 2x^2 +5x- 7x +2+1 = 0$

$(4x^2-2x^2) +(5x -7x) +(2+1) = 0$

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

We know that the quadratic formula to solve this,

X has two values which are $=\frac{(-b+\sqrt{b^{2}-4 a c})}{2 a} \text { and } \frac{-b-\sqrt{b^{2}-4 a c}}{2 a}$

Here a= 2; b = -2; c = 3

So substituting the values we get,

$x=\frac{-(-2)+\sqrt{(-2)^{2}-4 \times 2 \times 3}}{2 \times 2}$

$\Rightarrow\frac{2+\sqrt{4-24}}{4}=\frac{2+\sqrt{-20}}{4}=\frac{2+\sqrt{(-1) \times 4 \times 5}}{4}$

$\Rightarrow=\frac{2+2 i \sqrt{5}}{2}$   $(\text { assuming }-\sqrt{-1}=\mathrm{i})$

Again $x=\frac{2-2 i \sqrt{5}}{2}$

So, the roots are $\frac{2+2 i \sqrt{5}}{2}$ and $\frac{2-2 i \sqrt{5}}{2}$