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# help

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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
+252
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5

use calc

Apr 5, 2020
#2
+631
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Use quadratic equation. Specifically, the determinant.

$$b^2-4ac$$

25 - 4c = c

25 = 5c

c = 5

edited by AnExtremelyLongName  Apr 5, 2020
#3
+46
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$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$$
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
+631
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#5
+46
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Thanks!

derpiiXD  Apr 5, 2020
#6
+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