# How do you use the discriminant to determine the numbers of solutions of the quadratic equation 4x^2-20x + 25 = 0 and whether the solutions are real or complex?

Jun 18, 2018

one real root

root is $\frac{5}{2}$

#### Explanation:

Quadratic equations will have one or two roots (either real or complex roots)

Let our quadratic equation be $a {x}^{2} + b x + c = 0$

And ${x}_{1} , {x}_{2}$ be the roots of the equation

Then the discriminant of this quadratic equation will be ${b}^{2} - 4 a c$

$\textcolor{red}{1.}$ when ${b}^{2} - 4 a c > 0$ we will have two different real roots

${x}_{1} \mathmr{and} {x}_{2}$are real and unequal $\left({x}_{1} \ne {x}_{2}\right)$

$\textcolor{red}{2.}$ when ${b}^{2} - 4 a c = 0$ we will have two equal real roots

${x}_{1} \mathmr{and} {x}_{2}$ are real and ${x}_{1} = {x}_{2}$

$\textcolor{red}{3.}$when ${b}^{2} - 4 a c < 0$ we will have two complex roots

${x}_{1} \ne {x}_{2}$
${x}_{1} \mathmr{and} {x}_{2}$ are complex roots

we get the values of $a , b , c$ as $a = 4 , b = - 20 , c = 25$

The Discriminant will be ${\left(- 20\right)}^{2} - 4 \times 4 \times 25 \implies 400 - 400 \implies 0$

the roots are real and equal (Only one root)

Hence the equation must be a perfect square

$4 {x}^{2} - 20 x + 25 \implies {\left(2 x - 5\right)}^{2}$

Hence the root of the equation is $2 x - 5 = 0 \implies x = \frac{5}{2}$

${x}_{1} = {x}_{2} = \frac{5}{2}$