How do you solve this using the quadratic formula, $x^{2} + 10 x = - 25$ $x ^ { 2} + 10x = - 25$ ?

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Answer 1 · 2017-11-18T01:26:36

x = - 5

$x^{2} + 10 x + 25 = {(x + 5)}^{2}$ $x^2 + 10x + 25 = (x + 5)^2$
There is a double root --> x = - 5

Answer 2 · 2017-11-18T03:50:36

$x = - 5$ $x=-5$

$x^{2} + 10 x = - 25$ $x^2+10x=-25$

$x^{2} + 10 x + 25 = 0$ $x^2+10x+25=0$

The general form of a quadratic equation is $a x^{2} + b x + c$ $ax^2+bx+c$

Comparing, we get

$a = 1$ $a=1$
$b = 10$ $b=10$
$c = 25$ $c=25$
The quadratic formula is, $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

Plug in the values of $a, b$ $a,b$ and $c$ $c$

Therefore,

$x = \frac{- 10 \pm \sqrt{{(10)}^{2} - 4 \times 1 \times 25}}{2 \times 1}$ $x=(-10+-sqrt((10)^2-4 xx 1 xx 25))/(2 xx 1)$

$x = \frac{- 10 \pm \sqrt{100 - 100}}{2}$ $x=(-10+-sqrt(100-100))/2$

$x = \frac{- 10 \pm \sqrt{0}}{2}$ $x=(-10+-sqrt0)/2$

$x = \frac{- 10}{2}$ $x=(-10)/2$

$x = - 5$ $x=-5$

As the discriminant was equal to $0$ $0$ , there is only one root of this quadratic equation.

How do you solve this using the quadratic formula, x2+10x=−25x ^ { 2} + 10x = - 25?