How do you solve $x^{2} - 49 = 0$ $x^2 - 49 = 0$ ?

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Answer 1 · 2016-03-17T15:59:41

$x_{1, 2} = \pm 7$ $x_(1,2)=+-7$

You could use several methods.

The fastest is:

$x^{2} = 49$ $x^2=49$

$:.sqrt(x^2)=+-sqrt(49)$

$x_(1,2)=+-7$

I'd prefer to factorize the equation using the Square Difference Identitiy:

$(a^2-b^2)=(a+b)(a-b)$

to rewrite the second degree equation as product of first degree equation.

In your exercise:

$(x^2-49)=0 <=>(x^2-7^2)=0 <=>(x+7)(x-7)=0$

A product is zero when the factors are zero

$:.(x+7)=0=>x_1=-7$
$:.(x-7)=0=>x_2=7$

Therefore:

$x_(1,2)=+-7$

How do you solve x^2 - 49 = 0 x2−49=0x^2 - 49 = 0 ?