Solving Equations by Factoring: Solving the problem

We will factor using the shortened multiplication formulas:

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

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

Let's remember that there might be more than one solution for the value of x.

According to one solution, we'll take the square root:

$(x-5)^2=0$

$x-5=0$

$x=5$

According to the second solution, we'll use the shortened multiplication formula:

$(x-5)^2=x^2-10x+25=0$

We'll use the trinomial:

$(x-5)(x-5)=0$

$x-5=0$

$x=5$

or

$x-5=0$

$x=5$

Therefore, according to all calculations, $x=5$

We will factor using the shortened multiplication formulas:

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

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

Let's remember that there might be more than one solution for the value of x.

According to the first formula:

$x^2=a^2$

We'll take the square root:

$x=a$

$25=b^2$

We'll take the square root:

$b=5$

We'll use the first shortened multiplication formula:

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

$x^2-25=(x-5)(x+5)=0$

Therefore:

$x+5=0$

$x=-5$

Or:

$x-5=0$

$x=5$

We will factor using trinomials, remembering that there is more than one solution for the value of X:

$2x^2-7x+5=0$

We will factor -7X into two numbers whose product is 10:

$2x^2-5x-2x+5=0$

We will factor out a common factor:

$2x(x-1)-5(x-1)=0$

$(2x-5)(x-1)=0$

Therefore:

$x-1=0$$x=1$

Or:

$2x-5=0$

$2x=5$

$x=2.5$

First, factor using trinomials and remember that there might be more than one solution for the value of $x$:

$-x^2-7x-12=0$

Divide by -1:

$x^2+7x+12=0$

$(x+3)(x+4)=0$

Therefore:

$x+4=0$

$x=-4$

Or:

$x+3=0$

$x=-3$

Let's open the parentheses, remembering that there might be more than one solution for the value of X:

$(x-4)^2+x(x-12)=16$

$x^2-8x+16+x^2-12x=16$

$2x^2-20x=0$

$2x(x-10)=0$

Therefore:

$x-10=0$

$x=10$

Or:

$2x=0$

$x=0$