How do you solve $x^2 + 10x + 25 = 0$ using the quadratic formula?

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Answer 1 · 2017-05-22T02:13:23

$x = -5$

$x = (-b+-sqrt(b^2-4ac))/(2a)$

Where $ax^2+bx+c=0$ .

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In this case, $a = 1$ , $b=10$ , and $c=25$ . Now all we have to do is plug these values into the quadratic formula and simplify it.

$x = (-b+-sqrt(b^2-4ac))/(2a)$

$x = (-10+-sqrt(10^2-4(1)(25)))/(2(1))$

$x = (-10+-sqrt(100-100))/2$

$x = (-10+-0)/2$

Since adding or subtracting $0$ wouldn't change the answer at all, we actually only have a single solution instead of 2 solutions.

$x = (-10)/2 = -5$

Final Answer

Answer 2 · 2017-05-22T04:21:53

Double root at x = - 5

$y = x^2 + 10x + 25 = 0$
$D = b^2 - 4ac = 100 - 100 = 0$
Since D = 0, there is a double root at:
$x = -b/(2a) = -10/2 = - 5$

How do you solve x^2 + 10x + 25 = 0x^2 + 10x + 25 = 0 using the quadratic formula?