Question

What is the solution of the following equation 2x^2-3x+2=0

Answer 1

Answer:

$x= \frac{3+\sqrt{7}i}{4}}$ or $x=\frac{3-\sqrt{7}i}{4}}$

Step-by-step explanation:

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

We complete the square as follows

Group the constant term on the right hand side

$2x^2-3x=-2$

Divide through by 2,

$x^2-\frac{3}{2}x=-1$

Add half the square of the coefficient of x to get;

$x^2-\frac{3}{2}x+(-\frac{3}{4})^2=-1+(-\frac{3}{4})^2$

Simplify to get

$(x-\frac{3}{4})^2=-1+\frac{9}{16}$

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

Take the square root of both sides to get

$(x-\frac{3}{4})=\pm \sqrt{-\frac{7}{16}}$

Solve for x

$(x-\frac{3}{4})=\pm \frac{\sqrt{7}i}{4}}$

$x=\frac{3}{4} \pm \frac{\sqrt{7}i}{4}}$

$x= \frac{3+\sqrt{7}i}{4}}$ or $x=\frac{3-\sqrt{7}i}{4}}$