For what values of 'b' will the equation x^2 + bx + 81 = 0 have two equal roots?
A) -18 < b < 18
B) b < 18
C) b = + or - 18
D) b > 18
2007-10-12 19:26:31 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Use the formula to calculate the roots:
If your equation is ax^2 + bx +c = 0,
delta = b^2 - 4ac
In your case, b^2 -4ac = b^2 - 4*81. Set this equal to zero, since you want the roots to be equal and you get:
b^2 = 4*81. From this you can get b.
2007-10-12 19:33:42 · answer #1 · answered by mad_scientist 2 · 0⤊ 0⤋
x^2 + bx + 81 = 0
Since the equation have two equal roots,
b^2 - 4ac = 0
(b)^2 - 4(1)(81) = 0
b^2 - 324 = 0
(b + 18)(b - 18) = 0
b = -18 or b = 18
Therefore, the answer is C) b = ±18
2007-10-12 19:30:26 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Say you're given x^2+bx. How do you complete the square?
You take b/2 and square it. That makes the last term.
So 81=(b/2)^2
+/-9=b/2
2007-10-12 19:31:13 · answer #3 · answered by Amelia 6 · 0⤊ 0⤋
x = [- b ± â(b² - 324)] / 2
For equal roots, b² = 324
b = ± 18
2007-10-12 20:19:28 · answer #4 · answered by Como 7 · 0⤊ 0⤋