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

Question 732670: Solve the equation by completing the square t^2+t-28=0
Found 2 solutions by solver91311, MathLover1:
Answer by solver91311(24713) (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

Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it

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

here is your solution, you just put instead of and instead of :

Solved by pluggable solver: Completing the Square to Get a Quadratic into Vertex Form

Start with the given equation

Add to both sides

Factor out the leading coefficient

Take half of the x coefficient to get (ie ).

Now square to get (ie )

Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of does not change the equation

Now factor to get

Distribute

Multiply

Now add to both sides to isolate y

Combine like terms

Now the quadratic is in vertex form where , , and . Remember (h,k) is the vertex and "a" is the stretch/compression factor.

Check:

Notice if we graph the original equation we get:

Graph of . Notice how the vertex is (,).

Notice if we graph the final equation we get:

Graph of . Notice how the vertex is also (,).

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

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