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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!

Denzel  Aug 30, 2017
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#1
+76038
+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
#2
+25
+1

Tough to understand, could you give it with the looking of a had-written exercise, in a notebook?

Denzel  Aug 31, 2017
#3
+1088
+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
+1088
+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

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