SOLUTION: Good Morning Tutor, Could you tell me if I have these problems correct? Solve: x^2 + 7x - 1 = 0 is (-6 +- sqrt of 50 / 2) Solve by completeing the square: 2x

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Question 105112: Good Morning Tutor,
Could you tell me if I have these problems correct?
Solve:
x^2 + 7x - 1 = 0 is (-6 +- sqrt of 50 / 2)
Solve by completeing the square:
2x^2 + 8x - 5 = 0 is (4 +- sqrt of 26 / 2)
Thank you for your help this morning.

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

this is how you need to do it:
1.
Solved by pluggable solver: Quadratic Formula
Let's use the quadratic formula to solve for x:


Starting with the general quadratic





the general solution using the quadratic equation is:







So lets solve ( notice , , and )





Plug in a=1, b=7, and c=-1




Square 7 to get 49




Multiply to get




Combine like terms in the radicand (everything under the square root)




Simplify the square root (note: If you need help with simplifying the square root, check out this solver)




Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


or



Now break up the fraction



or



Simplify



or



So the solutions are:

or





2.
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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