How do you solve $(x-3)(x+5)=x^2-7x-15$ ?

Question

2053 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-27T14:50:06

$color(blue)(x=0$

$color(blue)((x-3)(x+5) )= x^2-7x-15$

As per property :

$color(blue)((x+a)(x+b) = x^2 + (a+b)x +ab$

Applying the same to the L.H.S of the expression ,
where: $a=-3$ and $b =5$ :

$color(blue)((x-3)(x+5) ) = x^2 + (-3 +5) x + (-3) xx5$

$=color(blue)(x^2 +2x -15$

The expression now becomes:

$color(blue)(x^2 +2x -15)= x^2-7x-15$

$cancel(color(blue)(x^2)) +2x -cancel15= cancelx^2-7x-cancel15$

$2x=-7x$

$2x +7x =0$

$9x =0$

$x =0/9$

$color(blue)(x=0$

How do you solve (x-3)(x+5)=x^2-7x-15(x-3)(x+5)=x^2-7x-15?