SOLUTION: solve the equation by completing the square. t^2-12t+20=0. how do i solve it?

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Question 124353: solve the equation by completing the square. t^2-12t+20=0. how do i solve it?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
note: I'm going to use x instead of t




y=x%5E2-12+x%2B20 Start with the given equation


y-20=x%5E2-12+x Subtract 20 from both sides


Take half of the x coefficient -12 to get -6 (ie %281%2F2%29%28-12%29=-6).

Now square -6 to get 36 (ie %28-6%29%5E2=%28-6%29%28-6%29=36)




y-20=x%5E2-12x%2B36-36 Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of 36 does not change the equation



y-20=%28x-6%29%5E2-36 Now factor x%5E2-12x%2B36 to get %28x-6%29%5E2



y=%28x-6%29%5E2-36%2B20 Now add 20 to both sides to isolate y


y=%28x-6%29%5E2-16 Combine like terms


Now we're done with completing the square


%28x-6%29%5E2-16=0 Now to solve for x, let y=0


%28x-6%29%5E2=%2B16 Add 16 to both sides



%28x-6%29%5E2=16 Reduce


Take the square root of both sides


Simplify the square root


Add 6 to both sides


So the solution breaks down to


x=6%2B4 or x=6-4

Combine like terms

x=10 or x=2


So our answers are


x=10 or x=2



Notice if we graph y=x%5E2-12+x%2B20, we can see that the roots are x=10 and x=2. So this visually verifies our answer.

graph%28500%2C500%2C-10%2C12%2C-10%2C12%2C1x%5E2-12x%2B20%29