# How do you find the value of the discriminant and state the type of solutions given 4a^2=8a-4?

##### 1 Answer
Sep 15, 2017

$\text{roots are real and equal}$

#### Explanation:

$\text{rearrange the quadratic equating to zero}$

$\Rightarrow 4 {a}^{2} - 8 a + 4 = 0 \leftarrow \textcolor{b l u e}{\text{ in standard form}}$

$\text{with } a = 4 , b = - 8 , c = 4$

$\Delta = {b}^{2} - 4 a c = {\left(- 8\right)}^{2} - \left(4 \times 4 \times 4\right)$

$\textcolor{w h i t e}{\times \times \times \times x} = 64 - 64 = 0$

$\textcolor{b l u e}{\text{Types of solutions}}$

• " if "Delta>0" roots are real and different"

• " if "Delta=0" roots are real and equal"

• " if "Delta<0" roots are not real"

$\text{here "Delta=0rArr" roots are real and equal}$

$\textcolor{b l u e}{\text{As a check}}$

$\text{solving } 4 {a}^{2} - 8 a + 4 = 0$

$\Rightarrow 4 \left({a}^{2} - 2 a + 1\right) = 0$

$\Rightarrow 4 \left(a - 1\right) \left(a - 1\right) = 0$

$\Rightarrow a = 1 \text{ or "a=1larrcolor(red)" real and equal}$