How do you find the roots, real and imaginary, of y=4x^2+12x-25 using the quadratic formula?

1 Answer
Mar 7, 2018

x=(-3+-sqrt34)/2

Explanation:

4x^2+12x-25=ax^2+bx+c

set 4x^2+12x-25 to 0 to find the roots real and imaginary

Using the discriminant b^2-4ac to find out if there are imaginary roots

(12)^2-4(4)(-25)=544

Delta("the discriminant")>0 So that means it have 2 unique real roots

Using the quadratic formula

x=(-(12)+-sqrt((12)^2-4(4)(-25)))/(2(4))

x=(-12+-sqrt544)/8

simplify the radical using root(n)xy=root(n)x*root(n)y

x=(-12+-sqrt(16*34))/8

x=(-12+-sqrt(4)sqrt(34))/8

x=(-12+-2sqrt(34))/8

Find the greatest common factor

x=(-3+-sqrt(34))/2

x=(-3+sqrt(34))/2 and -(3+sqrt(34))/2

Using C.T.S. (complete the square)
To get x^2 by itself divide the whole equation by 4

(4x^2+12x-25)/4=0/4

(cancel4x^2)/cancel4+(3cancel12x)/(1cancel4)-25/4=0

x^2+3xcancel(-25/4+25/4)=0+25/4

x^2+3x=25/4

complete the square by dividing the coefficient of x by 2 and then squaring it

(3/2)^2=9/4

add this term to both sides

x^2+3x+9/4=25/4+9/4

sqrt((x+3/2)^2)=+-sqrt(34/4

xcancel(+3/2-3/2)=(+-sqrt(34))/2+(-3)/2

x=(-3+-sqrt34)/2

:. x=(-3+sqrt34)/2 and (-3-sqrt34)/2