Question 648510
Original equations: {{{x^2 - x - 6 = 0}}} and {{{(x-8)2-9 = 0}}}

The zero-product principle states that the factored version of the equation must be set equal to 0 and solved for...

Let's factor and solve the first equation:

Step 1: Factor

{{{x^2 - x - 6 = 0}}} can be factored into {{{(x-3)(x+2) = 0}}}

Step 2: Solve each parenthesis individually

{{{(x-3)(x+2) = 0}}}
{{{(x-3)=0}}} and {{{(x+2) = 0}}}

You want to get 'x' by itself, so add '3' to both sides...

{{{(x-3)=0}}}
{{{x=3}}}

You want to get 'x' by itself, so subtract '2' to both sides...

{{{(x+2)=0}}}
{{{x=-2}}}

So... the roots of the equation {{{x^2 - x - 6 = 0}}} are {{{x=3}}} and {{{x=-2}}}

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Let's solve the other equation:

{{{(x-8)2-9 = 0}}}

Step 1: Let's rewrite the equation so that the '2' is placed in front of the parenthesis

{{{2(x-8)-9 = 0}}}

Step 2: Distribute the '2' to everything in the parenthesis

{{{2(x-8)-9 = 0}}}
{{{2x - 16 - 9 = 0}}}

Step 3: Combine like terms

{{{2x - 16 - 9 = 0}}}
{{{2x - 25 = 0}}}

Step 4: Add '25' to both sides in order to get the '2x' by itself

{{{2x - 25 = 0}}}
{{{2x = 25}}}

Step 5: Divide both sides by '2' in order to get 'x' by itself

{{{2x = 25}}}
{{{x = (25/2)}}}

The root of the equation {{{(x-8)2-9 = 0}}} is {{{x = (25/2)}}}