Question

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

4x2+5x + 2 = 2x2+7x-1

Answer 1

The roots of
`4 x^(2)+5 x+2=2 x^(2)+7 x-1` are
`(2+2 i √(5))/(2)` and
`(2-2 i √(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 * 2 * 3}}{2 * 2}`

`\Rightarrow(2+√(4-24))/(4)=(2+√(-20))/(4)=(2+√((-1) * 4 * 5))/(4)`

`\Rightarrow=(2+2 i √(5))/(2)`
`(\text { assuming }-√(-1)=\mathrm{i})`

Again
`x=(2-2 i √(5))/(2)`

So, the roots are
`(2+2 i √(5))/(2)` and
`(2-2 i √(5))/(2)`