SOLUTION: Solve the equation by completing the square t^2+t-28=0

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Question 732670: Solve the equation by completing the square t^2+t-28=0
Found 2 solutions by solver91311, MathLover1:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Step 1: Divide by the lead coefficient. Since the lead coefficient is 1, skip this step.

Step 2: Add the additive inverse of the constant term to both sides.

Step 3: Divide the 1st degree term coefficient by 2.

Step 4: Square the result of step 3.

Step 5: Add the result of both step 3 to both sides and collect like terms in the RHS.

Step 6: Factor the perfect square trinomial in the LHS.

Step 7: Take the square root of both sides, accounting for both the positive and negative roots by use of the symbol in the LHS

Step 8: Add the additive inverse of the constant term that remains in the RHS to both sides.

Step 9: Simplify the RHS expression.

John

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Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

here is your solution, you just put t instead of x and 0 instead of y:
Solved by pluggable solver: Completing the Square to Get a Quadratic into Vertex Form


y=1+x%5E2%2B1+x-28 Start with the given equation



y%2B28=1+x%5E2%2B1+x Add 28 to both sides



y%2B28=1%28x%5E2%2B1x%29 Factor out the leading coefficient 1



Take half of the x coefficient 1 to get 1%2F2 (ie %281%2F2%29%281%29=1%2F2).


Now square 1%2F2 to get 1%2F4 (ie %281%2F2%29%5E2=%281%2F2%29%281%2F2%29=1%2F4)





y%2B28=1%28x%5E2%2B1x%2B1%2F4-1%2F4%29 Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of 1%2F4 does not change the equation




y%2B28=1%28%28x%2B1%2F2%29%5E2-1%2F4%29 Now factor x%5E2%2B1x%2B1%2F4 to get %28x%2B1%2F2%29%5E2



y%2B28=1%28x%2B1%2F2%29%5E2-1%281%2F4%29 Distribute



y%2B28=1%28x%2B1%2F2%29%5E2-1%2F4 Multiply



y=1%28x%2B1%2F2%29%5E2-1%2F4-28 Now add %2B28 to both sides to isolate y



y=1%28x%2B1%2F2%29%5E2-113%2F4 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=1, h=-1%2F2, and k=-113%2F4. Remember (h,k) is the vertex and "a" is the stretch/compression factor.




Check:


Notice if we graph the original equation y=1x%5E2%2B1x-28 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1x%5E2%2B1x-28%29 Graph of y=1x%5E2%2B1x-28. Notice how the vertex is (-1%2F2,-113%2F4).



Notice if we graph the final equation y=1%28x%2B1%2F2%29%5E2-113%2F4 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1%28x%2B1%2F2%29%5E2-113%2F4%29 Graph of y=1%28x%2B1%2F2%29%5E2-113%2F4. Notice how the vertex is also (-1%2F2,-113%2F4).



So if these two equations were graphed on the same coordinate plane, one would overlap another perfectly. So this visually verifies our answer.