How do I use the quadratic formula to solve this equation?!
x^2 + 8x + 52 = 0
2007-12-04 07:23:57 · 4 answers · asked by Babygurl91 1 in Science & Mathematics ➔ Mathematics
OK
-b +- sqrt(b^2 -4ac)/2a
-8 +- sqrt(64 - 4(1)(52)) / 2(1)
-8 +- sqrt(64 -208) /2
-8 +- sqrt(-144) /2
-8 +- (12i) /2
2(-4 +-6i)/2
-4 + 6i or -4-6i
Proof
(-4+6i)^2 +8(-4+6i) +52 = 0??
16 -48i -36 -32 +48i +52 = 0??
16 -68 +52 = 0??
68-68= 0??
0 = 0 YES!!
Hope that helps.
2007-12-04 07:32:40 · answer #1 · answered by pyz01 7 · 0⤊ 0⤋
x = ( -b - sqrt [b^2 - 4ac] ) / 2a
where in your quadratic
a=1
b=8
c=52
x = (-8plusorminus 12i)/2 = -4 plus or minus 6i
2007-12-04 15:34:11 · answer #2 · answered by answerING 6 · 0⤊ 0⤋
a = 1, b = 8 c = 52
x = (-b +/- sqrt(b^2 - 4ac))/2a
= (-8 +/- sqrt(64-4x52))/2
=(-8 +/- sqrt(-144))/2
=(-8 +/- 12i)/2
=-4 +/- 6i
2007-12-04 15:33:13 · answer #3 · answered by norman 7 · 0⤊ 0⤋
a=1
b=8
c=52
delta=b*b-4*a*c
=64-4*1*52
=64- 208
<0
no real roots
The imaginary roots are
x1=(-8-sqrt(delta) )/2
x1=(-8+sqrt(delta) )/2
2007-12-04 15:29:51 · answer #4 · answered by iyiogrenci 6 · 0⤊ 0⤋