We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

+0

# Incomplete 2nd Grade Equations

0
421
4

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!

Aug 30, 2017

### 4+0 Answers

#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   Aug 30, 2017
#2
+1

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

Denzel  Aug 31, 2017
#3
+3

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.

Aug 31, 2017
#4
+2

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.

Aug 31, 2017