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X^2+14x-51=0 I have to solve by completing the square

User Nalaka
by
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1 Answer

1 vote

Answer:

Explanation:

Alright man, I got you.

So here is step-by-step

To solve the equation x^2 + 14x - 51 = 0 by completing the square, follow these steps:

Step 1: Move the constant term to the right-hand side

x^2 + 14x = 51

Step 2: Take half of the coefficient of x, square it, and add it to both sides of the equation

To find half of the coefficient of x, divide it by 2:

(14 / 2) = 7

Then square 7:

7^2 = 49

Add 49 to both sides of the equation:

x^2 + 14x + 49 = 51 + 49

Simplifying the right-hand side:

x^2 + 14x + 49 = 100

Step 3: Factor the left-hand side as a perfect square

The left-hand side is now a perfect square trinomial, which can be factored as:

(x + 7)^2 = 100

Step 4: Take the square root of both sides of the equation

Taking the square root of both sides of the equation gives:

x + 7 = ±10

Step 5: Solve for x

Subtracting 7 from both sides of the equation gives:

x = -7 ± 10

Therefore, the solutions to the equation x^2 + 14x - 51 = 0 are:

x = -7 + 10 = 3

or

x = -7 - 10 = -17

User AceKYD
by
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