SOLUTION: can you help me solve the system algebraically y=2(x+3)^2-5 y=14x+17

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Question 529656: can you help me solve the system algebraically
y=2(x+3)^2-5
y=14x+17
Found 2 solutions by lwsshak3, MathTherapy:
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
can you help me solve the system algebraically
y=2(x+3)^2-5
raise to power
2(x^2+6x+9)-5
distribute
y=2x^2+12x+18-5
combine like terms
y=2x^2+12x+13
set to zero
2x^2+12x+13=0
Solve for x using following quadratic formula:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
a=2, b=12, c=13
ans:
x≈-1.42
or
x≈-4.58

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
can you help me solve the system algebraically
y=2(x+3)^2-5
y=14x+17

Since the 2 equations are equal to y, then we can say that:
2%28x+%2B+3%29%5E2+-+5+=+14x+%2B+17

2%28x%5E2+%2B+6x+%2B+9%29+-+5+=+14x+%2B+17

2x%5E2+%2B+12x+%2B+18+-+5+=+14x+%2B+17

2x%5E2+%2B+12x+%2B+13+=+14x+%2B+17

2x%5E2+%2B+12x+-+14x+%2B+13+-+17+=+0

2x%5E2+-+2x+-+4+=+0

2%28x%5E2+-+x+-+2%29+=+2%280%29

x%5E2+-+x+-+2+=+0

(x - 2)(x + 1) = 0

x = 2 or – 1

When x = 2, then:
y = 14(2) + 17, or 28 + 17, or y+=+45. Solution: highlight_green%28x+=+2_when_y+=45%29

When x = - 1, then:
y = 14(- 1) + 17, or - 14 + 17, or y+=+3. Solution: highlight_green%28x+=-+1_when_y+=3%29

You should be able to do the check to make sure that the solution pairs make the 2 equations true.

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