A. 1
B. 6
C. 12
D. 36
E. 48
could you maybe show me how you figure out which one it is?
thanx =]
2007-05-10 14:41:24 · 6 answers · asked by jorie715 2 in Science & Mathematics ➔ Mathematics
If the equation ax^2+bx+c = 0 has only one solution, then b^2-4ac MUST equal zero.
Substituting for a = 1, b = -12 and c = k from your equation, we get
(-12)^2 - 4*1*k = 0
=> 4k = 144
Therefore,
k = 36
The answer is D
2007-05-10 14:45:14 · answer #1 · answered by cosine 2 · 0⤊ 0⤋
one million)here a root skill the fee of x. So, 2x^2 + px - 15 = 0 x = -5 2 * (-5)^2 + p*-5 - 15=0 2*25 - 5p = 15 50 - 5p = 15 shifting words 50-15=5p 35=5p p=7 putting fee of p interior the 2d equation, 7(x^2+x)+ok=0 7x^2+7x+ok=0 via the quadratic formula, D=b^2-4ac on condition that there are equivalent roots then d=0 0=40 9-4*7*ok 0=40 9-28k 28k=40 9 ok= 40 9/28 ok=7/4
2016-12-11 06:06:30 · answer #2 · answered by kuelper 3 · 0⤊ 0⤋
D
If (x^2 - 12x + k) is a square, then it will only have one root with multiplicity 2. A square is of the form: (x+a)^2 = x^2 + 2ax + a^2.
This means at -12 = 2a, and thus a = -6.
Then: k = a^2 = 36.
2007-05-10 14:46:01 · answer #3 · answered by NSurveyor 4 · 0⤊ 0⤋
D. 36
Do by completing squares
x^2 - 12x + k = 0
(x -6)^2 = 0
x^2 -12x +36 = 0
so k = 36
2007-05-10 14:44:22 · answer #4 · answered by looikk 4 · 0⤊ 0⤋
Only one solution (ie. two equal solutions) --- the quadratic must have the form
x^2 + 2ax + a^2 = 0
2ax = -12x
a=-6
a^2 = 36
k should be 36
2007-05-10 14:46:01 · answer #5 · answered by gudspeling 7 · 0⤊ 0⤋
(x - 6).(x - 6) = x² - 12x + k = 0
x² - 12x + 36 = x² - 12x + k = 0
k = 36
ANSWER D
2007-05-11 06:45:09 · answer #6 · answered by Como 7 · 0⤊ 0⤋