How do you solve $x^2 = 14x – 49$ ?

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Answer 1 · 2015-07-08T08:25:45

The answer is $x = 7$ .

$x^2=14x-49$

Subtract $14x$ from both sides of the equation.

$x^2-14x=-49$

Add $49$ to both sides of the equation.

$x^2-14x+49=0$

$x^2-14x+49$ is a perfect square trinomial in the form $a^2+2ab+b^2=(a+b)^2$ , where $a=x and b=-7$ .

$x^2-14x+49=0$ =

$(x-7)^2=0$

Take the square root of both sides.

$(x-7)=sqrt 0$ =

$x-7=0$ =

$x=7$

How do you solve x^2 = 14x – 49x^2 = 14x – 49?