SOLUTION: y=2x^2,y=x+3 solve using system of equations algebraically.

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Question 449742: y=2x^2,y=x+3 solve using system of equations algebraically.
Answer by Math_Teacher_37(4) About Me  (Show Source):
You can put this solution on YOUR website!
+y+=+2x%5E2 and y+=+x+%2B+3
Since +y+=+2x%5E2 and y+=+x+%2B+3, this means that 2x%5E2+=+x+%2B+3.
Now, you have to solve for x, using the quadratic formula. Begin by moving the terms on the right side of the equation to the left side:
2x%5E2+-+x+-+3+=+0
Now, in the quadratic form, ax%5E2+%2B+bx+%2B+c, a = 2, b = -1, and c = -3. Substitute these values into the quadratic formula, as shown below.
x+=+%28-b+%2B-sqrt%28b%5E2-4%2Aa%2Ac%29%29%2F%282%2Aa%29

x+=+%28-%28-1%29+%2B-sqrt%28%28-1%29%5E2+-+4%2A2%2A-3%29%29%2F%282%2A2%29

x+=+%281+%2B-sqrt%281+%2B+24%29%29%2F%284%29

x+=+%281+%2B-sqrt%2825%29%29%2F%284%29

x+=+%281+%2B-+5%29%2F4

x+=+%281%2B5%29%2F4 OR x+=+%281-5%29%2F4

x+=+6%2F4 OR x+=+-4%2F4

x+=+3%2F2 OR x+=+-1

The solutions are x = -1 OR x = 3/2.