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?

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Answer 1 · 2018-06-18T10:37:37

one real root

root is $5/2$

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

Let our quadratic equation be $ax^2+bx+c=0$

And $x_1,x_2$ be the roots of the equation

Then the discriminant of this quadratic equation will be $b^2-4ac$

$color(red)1.$ when $b^2-4ac>0$ we will have two different real roots

$x_1 and x_2$ are real and unequal $(x_1!=x_2)$

$color(red)2.$ when $b^2-4ac=0$ we will have two equal real roots

$x_1 and x_2$ are real and $x_1=x_2$

$color(red)3.$ when $b^2-4ac<0$ we will have two complex roots

$x_1!=x_2$
$x_1 and x_2$ are complex roots

Comparing the given quadratic equation with the general quadratic equation

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

The Discriminant will be $(-20)^2-4xx4xx25=>400-400=>0$

the roots are real and equal (Only one root)

Hence the equation must be a perfect square

$4x^2-20x+25=>(2x-5)^2$

Hence the root of the equation is $2x-5=0=>x=5/2$

$x_1=x_2=5/2$

How do you use the discriminant to determine the numbers of solutions of the quadratic equation 4x^2-20x + 25 = 04x^2-20x + 25 = 0 and whether the solutions are real or complex?