Solve the equation . Show work.

(x-5)x=14

Solve the equation . Show work.

(x-5)x=14

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Westford, MA

(x-5)x = 14

Distribute x:

x^2 - 5x = 14

Add -14 to both sides:

1x^2 - 5x - 14 = 0

1(-14) = 2(-7) and 2 - 7 = -5:

x^2 + 2x - 7x - 14 = 0

x(x + 2) - 7(x + 2) = 0

(x + 2)(x - 7) = 0

Zero Product Property:

x + 2 = 0 or x - 7 = 0

x = -2 or x = 7

check:

((-2)-5)(-2) =? 14

(-7)(-2) = 14 √

((7)-5)(7) =? 14

(2)(7) = 14 √

West Richland, WA

For this one, since the amounts in the parenthesis are being multiplied by x, you distribute x to all of the terms inside the parenthesis. When you do this, the problem looks like this:

x^2-5x=14

Now we have a quadratic, so we need to set the equation to 0. Subtract both sides by 14:

x^2-5x-14=0

We can then factor the trinomial:

(x-7)(x+2)=0

Now we just need to solve for x so that when we plug it into the equation, it will equal zero. We find that in the first binomial, if x=7 and we multiply it by whatever is in the second binomial, the answer is zero. By the same method, if x=-2 and we multiply that by whatever is in the first binomial, the answer is zero. This means that x=7 and x=-2.

x^2-5x=14

Now we have a quadratic, so we need to set the equation to 0. Subtract both sides by 14:

x^2-5x-14=0

We can then factor the trinomial:

(x-7)(x+2)=0

Now we just need to solve for x so that when we plug it into the equation, it will equal zero. We find that in the first binomial, if x=7 and we multiply it by whatever is in the second binomial, the answer is zero. By the same method, if x=-2 and we multiply that by whatever is in the first binomial, the answer is zero. This means that x=7 and x=-2.

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