SOLUTION: Solve each of the following quadratic equations by completing the square. sect:10.2 # 22 x^2-6x-3=0 Section 10.2 # 28 2x^2+10x+11=0 Sect. 10.2 # 38 Number problems. F

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Solve each of the following quadratic equations by completing the square. sect:10.2 # 22 x^2-6x-3=0 Section 10.2 # 28 2x^2+10x+11=0 Sect. 10.2 # 38 Number problems. F      Log On


   



Question 73989: Solve each of the following quadratic equations by completing the square.
sect:10.2 # 22
x^2-6x-3=0
Section 10.2 # 28
2x^2+10x+11=0
Sect. 10.2 # 38
Number problems. Find two consecutive positive integers such that the sum of
their squares is 85.


Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Given: x^2-6x-3=0
.
Solve by completing the square
.
The first thing to do is to make sure the coefficient (the multiplier) of the x^2 term is 1.
In this case it is. So we can move on to the next step. Use parentheses to separate the
two terms that contain x from the rest of the equation. This becomes:
.
(x^2 - 6x )-3 = 0
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Divide the coefficient of the x term by 2. In this case it is -6/2 = -3. Square this result
to get (-3)^2 = 9. Add 9 inside the parentheses, but in order to do this without changing the
equation you need to subtract 9 outside the parentheses. This makes the equation become:
.
(x^2 - 6x + 9) - 3 - 9 = 0
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Combine the two numbers outside the parentheses to get:
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(x^2 - 6x + 9) - 12 = 0
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Add +12 to both sides to eliminate the -12 on the left side and get:
.
(x^2 - 6x + 9) = 12
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Note that the trinomial in the parentheses is a perfect square. It can be written as
(x - 3)^2. Substitute this into the equation and you get:
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(x-3)^2 = 12
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Now take the square root of both sides. I'm going to use +- to mean plus and minus. The
square root of both sides results in:
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x-3 = +- sqrt(12)
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Eliminate the -3 on the left side by adding 3 to both sides to get:
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x = 3 +- sqrt(12)
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As a final simplification note that sqrt(12) = sqrt(4*3) = sqrt(4)* sqrt(3) = 2*sqrt(3).
Substitute this for sqrt(12) and the answer becomes:
.
x = 3 +- 2*sqrt(3)
.
On to the next problem: solve by completing the squares: 2x^2+10x+11=0
.
Recall that I mentioned you want the coefficient of the x^2 term to be 1. In this case
it is not. So divide the entire equation (all terms on both sides by 2 to reduce it to:
.
x^2 + 5x + 11/2 = 0
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Convert the 11/2 term to 5.5 by dividing 2 into 11.
.
Again use parentheses to isolate the terms containing x and get:
.
(x^2 + 5x ) + 5.5 = 0
.
Take half of the coefficient of the x term and square it. Half of 5 is 2.5 which squares
to 6.25. Add it inside the parentheses and subtract it outside the parentheses to get:
.
(x^2 + 5x + 6.25) + 5.5 - 6.25 = 0
.
Combine the two constants outside the parentheses to get:
.
(x^2 + 5x + 6.25) - 0.75 = 0
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Eliminate the - 0.75 from the left side by adding 0.75 to both sides and you have:
.
(x^2 + 5x + 6.25) = 0.75
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The term in the parentheses is a perfect square so it can be written as such and the equation becomes:
.
(x + 2.5)^2 = 0.75
.
Take the square root of both sides and the result is:
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x+2.5 = +- sqrt(0.75)
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Eliminate the 2.5 on the left side by subtracting 2.5 from both sides to get:
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x = -2.5 +- sqrt(0.75)
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Note that the sqrt(0.75) = sqrt(0.25*3) = sqrt(0.25)*sqrt(3) = 0.5*sqrt(3)
.
Substitute this and the answer becomes:
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x = -2.5 +- 0.5*sqrt(3)
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And on to the final problem. Let one of the integers be N. Then the next consecutive
integer is N + 1. The sum of their squares can be written as:
.
N^2 + (N+1)^2
.
and the problem says the sum of these two squares equals 85. So the equation becomes:
.
N^2 + (N+1)^2 = 85
.
Square the term in parentheses on the left side to get:
.
N^2 + N^2 + 2N + 1 = 85
.
Combine the N^2 terms and the equation becomes:
.
2N^2 + 2N + 1 = 85
.
To make things a little easier let's eliminate the +1 on the left side by adding -1 to both
sides to get:
.
2N^2 + 2N = 84
.
We need a coefficient of 1 on the N^2 term so divide the entire equation (all terms on
both sides by 2 to get:
.
N^2 + N = 42
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Just noticed something here. We don't need to solve this one by completing the square.
Subtract 42 from both sides to get:
.
N^2 + N - 42 = 0
.
This can be factored into:
.
(N+7)*(N-6) = 0
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And this equation will be true if either N = -7 or N = +6 because those are the two values
that make one of the factors equal to zero. The problem limits the answer to positive
integers so we can forget N = -7. The two integers are +6 and the next consecutive
integer of +7.
.
Check.
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6^2 + 7^2 = 36 + 49 = 85 . Yup. It works.
If you must work it by completing the square you can use the same process as in the preceding
two problems, and you know that you should get an answer of 6 for N.
.
Hope this helps you to understand and solve these types of problems.