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# Module 16

### Module 16 - Solving Quadratic Equations by Completing the Square

Question | Answer |
---|---|

Solve the equation by completing the square x^2+10x+22=0 | To solve a quadratic equation by completing the square add a constant to both sides of the equation so that the remaining trinomial is a perfect square trinomial. |

The coefficient of x^2 term of the quadratic equation must be equal to 1 in order to determine the constant to be added to both sides of the equation. Is the leading coefficient equal to 1? x^2+10x+22=0 yes or no? | If you answered yes, you are correct!! :)) |

Rewrite the equation with the constant by itself on the right side of the equation. x^2+10x+22=0 | Subtract 22 from both sides X^2+10x=-22 |

Now take 1/2 of the numerical coefficient of the x-term and square it. x term is equal to 10x | 1/2* (10) = 5 now square it (5)^2 = 25 |

Add the constant 25 to both sides of the equation to form a perfect square trinomial on the left side of the equation as the square of a binomial x^2+10x=-22 | Now add 25 to both sides... x^2=10x=25=-22+25 |

Now factor the left side x^2=10x=25=-22+25 | It should look like this: (x+5)^2= -22+25 |

Now, simplify the right side of the equation (x+5)^2= -22+25 | Which should look like this: (x+5)^2= 3 |

The square root property is stated as.... | If x^2=a where a is a real number then x=+/-√a |

Use the square root property. Remember that the value on the right can be positive or negative. Solve for x. | (x+5)^2 = 3 x+5+=/-√3 x+5=√3 or x+5= -√3 x= -5+√3, -5-√3 Congrats you made it through!! :) |

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