+27 How much does c value if the equation 2x2-8x +c=0 must have two solutions with the same value
3x2+bx+12=0 b? The equatin must have two same solutions
x2+bx-24=0 b? The solution must be -3
x2-mx+8=0 m? The solution must be 4
write an equation with solutions -2 and -17
write an equation with solutions -11 and 9, and the coefficient of x2 must he -4
Hope you guys can give me some help. Thxs!
+128474 3x^2+bx+12=0 b? The equation must have two same solutions
The discriminant must = 0 if this is true
So b^2 - 4(3)(12) = 0
b^2 = 144
b = ±12 [ -2 and 2 will be solutions.....each with a multiplicity of 2 ]
x^2+bx-24=0 b? The solution must be -3
We must have this
(-3)^2 + b(-3) - 24 = 0
9 - 3b - 24 = 0
-15 - 3b = 0
-15 = 3b
b = -5 [ 8 will also be a solution ]
x^2-mx+8=0 m? The solution must be 4
(4)^2 - m(4) + 8 = 0
16 - 4m + 8 = 0
24 - 4m = 0
24 = 4m
6 = m [ 2 will also be a solution ]
Write an equation with solutions -2 and -17
P(x) = (x + 2) (x + `17) = x^2 + 19x + 34
Write an equation with solutions -11 and 9, and the coefficient of x^2 must be -4
We have
P(x) = -4 (x + 11) (x - 9) = -4 [x^2 + 2x - 99] = -4x^2 - 8x + 396
+2441 It does not appear as if Cphill answered the first question of for what value of c makes \(2x^2+8x+c=0\) have only 1 unique solution.
At first, I was not sure how to approach it, but then I remembered something called the discriminant. The discriminant, in a quadratic equation \(ax^2+bx+c\) is \(b^2-4ac\). The discriminant can tell you a few things about the number of solutions in a quadratic.
1) If \(b^2-4ac>0\), then there are 2 unique solutions
2) If \(b^2-4ac<0\), then there are no real solutions
3) If \(b^2-4ac=0\), then there is 1 unique solution
Based on the rules I had just described, we should definitely use condition #3 because we want the quadratic to have 1 unique solution.
| \(b^2-4ac=0\) | Plug in the values we know. |
| \((-8)^2-4(2)(c)=0\) | Let's simplify the right hand side. |
| \(64-8c=0\) | Subtract 64 from both sides of the equation. |
| \(-8c=-64\) | Divide by -8 on both sides. |
| \(c=8\) | |
Therefore, \(c=8\) is the only value for c such that there is only one unique solution.
+2441 You said you had trouble understanding Cphill's work. Maybe this will help:
1.
3x^2+bx+12=0 b? The equation must have two same solutions
The discriminant, \(b^2-4ac\), must =0 if this is true
So \(b^2-4(3)(12)=0\)
\(b^2=144\)
\(b=\pm12\)
[ -2 and 2 will be solutions.....each with a multiplicity of 2 ]
2.
x^2+bx-24=0 b? The solution must be -3
We must have the following:
\((-3)^2 + b(-3) - 24 = 0\)
\(9 - 3b - 24 = 0\)
\(-15 - 3b = 0\)
\(-15 = 3b\)
\(b = -5 \)
[ 8 will also be a solution ]
3.
x^2-mx+8=0 m? The solution must be 4
\((4)^2 - m(4) + 8 = 0\)
\(16 - 4m + 8 = 0\)
\(24 - 4m = 0\)
\(24 = 4m\)
\(m=6\)
[ 2 will also be a solution ]
4.
Write an equation with solutions -2 and -17
\(P(x) = (x + 2) (x + 17) = x^2 + 19x + 34\)
5.
Write an equation with solutions -11 and 9, and the coefficient of x^2 must be -4
We have
\(P(x) = -4 (x + 11) (x - 9) = -4 [x^2 + 2x - 99] = -4x^2 - 8x + 396 \)
I hiope this helped.
