How do you find the discriminant, describe the number and type of root, and find the exact solution using the quadratic formula given $x^2-16x+4=0$ ?

Question

3188 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-30T01:48:47

$discriminant = sqrt(240)$

Equation has two real roots

Two roots are $color(red)(x = 23.746, 8.254)$

Given quadratic equation is $x^2 - 16x + 4 = 0$

Formula to find the roots $x = (-b +- sqrt(b^2 - (4 a c )) / (2a))$

Where a, b, c are the coefficients of $x^2, x, const.$ terms respectively.

$a = 1, b = -16, c = 4$

Term $sqrt(b^2 - (4ac))$ is called the discriminant.

If the discriminant term is

a) Positive - both the roots are real

b) Zero - one real solution

c) Negative - two complex solutions

$discriminant = sqrt((-16)^2 - (4 * 1 * 4)) = sqrt(240)$

Hence the equation has two real roots.

$x = -(-16) +- sqrt((-16)^2 - (4 * 1 * 4)) / (2 * 1)$

$x = (16 +- sqrt(256 - 16)) / 2 = 8 +- (sqrt240)/2$

$x = 16 +- ((cancel2 sqrt60)/cancel2)$

$x = 16 +- 7.746$

$color(red)(x = 23.746, 8.254)$

How do you find the discriminant, describe the number and type of root, and find the exact solution using the quadratic formula given x^2-16x+4=0x^2-16x+4=0?