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?
+658 Use quadratic equation. Specifically, the determinant.
\(b^2-4ac\)
25 - 4c = c
25 = 5c
c = 5
+46 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
+500 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:
\(c = 5\)
