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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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4+0 Answers

 #1
avatar+76160 
+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

 

 

cool cool cool

CPhill  Aug 30, 2017
 #2
avatar+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
avatar+1119 
+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
avatar+1119 
+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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