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#?

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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)#