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The quadratic x^2 + 5x + c has roots in the form of \(x = \frac{-5 \pm \sqrt{c}}{2}\). What is the value of c?

 Apr 5, 2020
 #1
avatar+252 
0

5

use calc

 

coolcoolcool

 Apr 5, 2020
 #2
avatar+631 
+1

Use quadratic equation. Specifically, the determinant.

\(b^2-4ac\)

 

25 - 4c = c

25 = 5c

c = 5

edited by AnExtremelyLongName  Apr 5, 2020
 #3
avatar+46 
+3

Use the quadratic formula:

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
x^2+5x+c=ax^2+bx+c

a=1

b=5

c=?

Plug it into the formula and we get

\(x = {-5 \pm \sqrt{5^2-4(1)c} \over 2(1)}\)

and solve to get

\(x = {-5 \pm \sqrt{25-4c} \over 2}\)

now, we know that everything is the same except for the discriminant(the stuff in the square root), and we also know that the disriminant is c

Set them equal to get

25-4c=c

Now solve to get

25=5c

and c=5

 

XD

 Apr 5, 2020
 #4
avatar+631 
0

I wish CPhill can nominate your answer as a "best" answer.

 #5
avatar+46 
+2

Thanks!

derpiiXD  Apr 5, 2020
 #6
avatar+485 
+1

Because we have the roots of this quadratic given as : 

 

\(x = {-5 \pm \sqrt{c} \over 2}\), by vietas, we can write:

\({-5 + \sqrt{c} \over 2} * {-5 - \sqrt{c} \over 2}\) = c

Expanding, we get:

\({25-c\over4} = c\), because the square roots cancle out

then, this gives us:

\(4c = 25-c\) when we multiply by 4 on both sides

\(5c = 25\)

Dividing by 5 on both sides, we get:

\(c = 5\)

.
 Apr 5, 2020

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