SOLUTION: Hello, I need some help on quadratic equations. Any and all help will be appreciated. 1. Solve using the quadratic formula: x^2 - 3x = 4x -1 2. Solve using the quadratic f

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Hello, I need some help on quadratic equations. Any and all help will be appreciated. 1. Solve using the quadratic formula: x^2 - 3x = 4x -1 2. Solve using the quadratic f      Log On


   



Question 202134: Hello,
I need some help on quadratic equations. Any and all help will be appreciated.
1. Solve using the quadratic formula: x^2 - 3x = 4x -1
2. Solve using the quadratic formula: x^2 -5x - 1 = -7
3. Solve by completing the square: x^2 + 8x + 2 = 0
Thank you in advance for your help.

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
the quadratic formula is:
x = %28-b+%2B-+sqrt%28b%5E2-4%2Aa%2Ac%29%29%2F%282%2Aa%29
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the general form of the quadratic equation is:
a%2Ax%5E2+%2B+b%2Ax+%2B+c+=+0
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to use the quadratic equation, you find the values for a, b, and c, and plug them into the quadratic equation and solve.
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your first problem is:
x%5E2+-+3x+=+4x+-1
first you need to convert it to the standard form.
subtract 4x from both sides of the equation and add 1 to both sides of the equation to get:
x%5E2+-+7x+%2B+1+=+0
a = 1
b = -7
c = 1
plug these values into the quadratic equation to get:
x = +%28-%28-7%29+%2B-+sqrt+%28%28-7%29%5E2+-+4%2A1%2A1%29%29%2F%282%2A1%29
which becomes:
x = +%287+%2B-+sqrt%2845%29%29%2F2%29+
this answer works as is but you can reduce it further by simplifying the number under the square root sign as follows:
x = +%28+7+%2B-+sqrt%283%2A3%2A5%29%29%2F2+
which reduces to:
x = +%28+7+%2B-+3%2Asqrt%285%29%29%2F2+
all you need to do now is verify the answer is correct by plugging those values for x into the original equation to see if the equation is true. I cheated by using my calculator, but the answers are verified to be true so these answers are good.
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your second problem is:
x^2 -5x - 1 = -7
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you do the same thing.
convert it to standard form.
get the values of a, b, and c.
plug them into the formulas and solve.
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i'll solve without explanation this time and you can follow along or try to solve it on your own.
x^2 - 5x - 1 = -7
add 7 to both sides to get:
x^2 - 5x + 6 = 0
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a = 1
b = -5
c = 6
x = %28-b+%2B-+sqrt%28b%5E2-4%2Aa%2Ac%29%29%2F%282%2Aa%29
x = %28-%28-5%29+%2B-+sqrt%28%28-5%29%5E2-4%2A1%2A6%29%29%2F%282%2A1%29
which becomes
x = %285+%2B-+sqrt%2825-24%29%29%2F2
which becomes
x = %285+%2B-+sqrt%281%29%29%2F2
which becomes
x = %285+%2B-+1%29%2F2
which becomes
x = 3 or x = 2
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plug these values into the original equations to see if the equations are true.
original equation is:
x^2 -5x - 1 = -7
when x = 2 this becomes
4 - 10 - 1 = -7
which becomes
-6-1 = -7 which is true.
when x = 3 this becomes
9 - 15 - 1 = -7
which becomes
-6 -1 = -7 which is also true.
the answers are good.
x = %285+%2B-+1%29%2F2
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your last equation needs to be solved by completing the squares.
this is a different technique but gets you to the same answer.
you could also use the quadratic equation here as well, but sometimes completing the square is easier.
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your last equation is:
x%5E2+%2B+8x+%2B+2+=+0+
you would subtract 2 from both sides of the equation to get:
x%5E2+%2B+8x+=+-2
you take 8/2 = 4 and use that value to complete the square.
you take %28x%2B4%29%5E2 to get:
x%5E2+%2B+8%2Ax+%2B+16.
so what you have is:
%28x%2B4%29%5E2+=+x%5E2+%2B+8%2Ax+%2B+16.
if you subtract 16 from that, you have:
x%5E2+%2B+8%2Ax+=%28x%2B4%29%5E2+-+16
you can now go back to your original equation after modification which was:
x%5E2+%2B+8%2Ax+=+-2
and substitute %28x%2B4%29%5E2+-+16 for x%5E2+%2B+8%2Ax to get:
%28x%2B4%29%5E2+-+16+=+-2
you add 16 to both sides of the equation to get:
%28x%2B4%29%5E2+=+14
you have just completed the square.
solving by taking the square root of both sides and you will get:
%28x+%2B+4%29 = +/- sqrt%2814%29
which becomes:
+x+=+-4+%2B-+sqrt%2814%29+
you verify that this answer is valid by plugging it into the original equation which was:
x%5E2+%2B+8%2Ax+%2B+2+=+0
the values check out ok so the answer is good.
note that if you had solved this equation using the quadratic formula, you would have gotten the same answer.
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note that the standard form of the quadratic equation is:
ax%5E2+%2B+bx+%2B+c.
to complete the square, you need to move the c over to the right side and you need to divide both sides of the equation by a, if a is not already 1.
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you will wind up with:
ax%5E2+%2B+bx+=+-c
before dividing by a if it is not 1 to start with.
if a is not one, you will wind up with:
%28ax%5E2+%2B+bx%29%2Fa+=+-c%2Fa
which is the same as:
x%5E2+%2B+%28bx%29%2Fa+=+-c%2Fa
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i'll show you how that works with a simple example.
once you have the standard form for completing the square, the operations are the same as above.
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take 3x%5E2+%2B+9x+%2B+6+=+0
this is the standard form of the quadratic equation of ax%5E2+%2B+bx+%2B+c.
note that a is not = to 1
first step:
move the c over to the right hand side by subtracting it from both sides of the equation.
you get:
3x%5E2+%2B+9x+=+-6
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since the a coefficient is not 1 (it's 3), you need to divide both sides of the equation by 3 to make it 1.
it becomes:
%283x%5E2+%2B+9x%29%2F3+=+-6%2F3
which becomes:
x%5E2+%2B+3x+=+-2
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now you're in the standard form for completing the square.
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take 1/2 the b factor which means take 1/2 of 3.
you get 3/2
your factor that will be squared is %28x+%2B+3%2F2%29.
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you get:
%28x+%2B+3%2F2%29%5E2+=+x%5E2+%2B+3x+%2B+%289%2F4%29
your left side of the equation was %28x%5E2+%2B+3x%29
this equals %28x+%2B+3%2F2%29%5E2+-+%289%2F4%29
your equation which was:
x%5E2+%2B+3x+=+-2 becomes:
%28x+%2B+3%2F2%29%5E2+-+%289%2F4%29+=+-2
you add (9/4) to both sides to get:
%28x+%2B+%283%2F2%29%29%5E2+=+-2+%2B+%289%2F4%29
this becomes:
%28x+%2B+%283%2F2%29%29%5E2+=+%28-8%2F4%29+%2B+%289%2F4%29
which becomes:
%28x+%2B+%283%2F2%29%29%5E2+=+%281%2F4%29
take the square root of both sides to get:
x+%2B+%283%2F2%29 = +/- sqrt%281%2F4%29
this means:
x+=+-%283%2F2%29+%2B+sqrt+%281%2F4%29
or
x+=+-%283%2F2%29+-+sqrt+%281%2F4%29
or
x = both.
you wind up with:
x = -1 or x = -2 or both.
both values of x checked out in the original equation so they're both good.
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