# Use the quadratic formula to solve 8x^2 -5x +2 =0.

## Question:

Use the quadratic formula to solve {eq}\, 8x^2 -5x +2 =0 {/eq}.

An equation which is of the form {eq}ax^2+bx+c=0 {/eq} where {eq}a, b \text{ and } c {/eq} are constants is called a quadratic equation in {eq}x {/eq}. It has two (either same or distinct) roots. The roots can be solved by using the quadratic formula which states:

$$x=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$$

The given equation is:

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

Comparing this with {eq}ax^2+bx+c=0 {/eq}, we get:

$$a=8 \\ b=-5\\ c=2$$

We substitute these values in the quadratic formula:

\begin{align} x& =\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\\[0.4cm] &= \frac{-(-5) \pm \sqrt{(-5)^{2}-4 \cdot 8 \cdot 2}}{2 \cdot 8} \\[0.4cm] &= \frac{5 \pm \sqrt{25-64}}{16} \\[0.4cm] &= \frac{5 \pm \sqrt{25-64}}{16} \\[0.4cm] &= \frac{5 \pm \sqrt{-39}}{16} \\[0.4cm] &= \frac{5 \pm i \sqrt{39}}{16}& [ \because \sqrt{-1}=i ] \\[0.4cm] &= \boxed{\mathbf{ \frac{5}{16}+i \frac{\sqrt{39}}{16}; \,\,\, \frac{5}{16}-i \frac{\sqrt{39}}{16}}} \end{align}

from Math 101: College Algebra

Chapter 4 / Lesson 10
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