How much does c value if the equation 2x^{2}-8x +c=0 must have two solutions with the same value

3x^{2}+bx+12=0 b? The equatin must have two same solutions

x^{2}+bx-24=0 b? The solution must be -3

x^{2}-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 x^{2} must he -4

Hope you guys can give me some help. Thxs!

Denzel
Aug 30, 2017

#1**+3 **

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

CPhill
Aug 30, 2017

#3**+2 **

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.

TheXSquaredFactor
Aug 31, 2017

#4**+1 **

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.

TheXSquaredFactor
Aug 31, 2017