SOLUTION: Hello! I need to find the range of a quadratic equation, which is 2x^2+ -2x + 3 = 0 Could you please help me??

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Question 949629: Hello! I need to find the range of a quadratic equation, which is 2x^2+ -2x + 3 = 0
Could you please help me??

Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39627) About Me  (Show Source):
You can put this solution on YOUR website!
Above the vertex. Find the minimum y value, "k", in the standard form equation.

2x%5E2%2B%28-2x%29%2B3=0
2x%5E2-2x%2B3=0
2%28x%5E2-x%2B3%2F2%29=0, ---------------- %281%2F2%29%5E2 will complete the square.
highlight_green%282%28x-1%2F2%29%5E2%2B5%2F2=0%29

Actually, "range of the quadratic equation" is wrong. The solution process just given
(although missing some arithmetic steps) is the function equated to 0, allowing you
to see the vertex of the function, which is read directly from the equation to be
(1/2, 5/2).

The EQUATION can be solved easily for the x values, which are the ROOTS or ZEROS of
the quadratic function. The RANGE of the function from which the equation comes is
highlight%285%2F2%3Cy%29.

Answer by MathTherapy(10556) About Me  (Show Source):
You can put this solution on YOUR website!

Hello! I need to find the range of a quadratic equation, which is 2x^2+ -2x + 3 = 0
Could you please help me??
2x%5E2+-+2x+%2B+3+=+0
This equation will have a minimum, since a > 0
This minimum will occur at: x+=+-+b%2F2a______x+=+-+-+2%2F%282%2A2%29_______x+=+2%2F4_______x+=+1%2F2
Minimum point: y+=+2%281%2F2%29%5E2+-+2%281%2F2%29+%2B+3____y+=+2%281%2F4%29+-+1+%2B+3_____y+=+1%2F2+-+1+%2B+3____y+=+2%261%2F2____y+=+5%2F4
Therefore, range is: highlight_green%285%2F4+%3C=+y+%3C+infinity%29