Author Question: Solving quadratics using vertex form/root form? (Read 1459 times)

Jones

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I've been trying really hard to figure out these for Precalc, but I just don't understand. How do I solve this quadratic using vertex form?

4x^2 + 149 = 28x

and this one, using root form?

3x + 8 + 16/3x = 0

thanks so much!
never mind the root form bit. apparently, that's just basic factoring.



Sandstorm

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You find the vertex of a quadratic function by completing the square.

4x^2 - 28x + 149 = y
4(x^2 - 7x) = y - 149
x^2 - 7x = (1 / 4)(y - 149)
x^2 - 7x + (7 / 2)^2 = (1 / 4)(y - 149 + 49)
= (1 / 4)(y - 100)
(x - 7 / 2)^2 = (1 / 4)(y - 100)

That's the vertex form of the equation, giving vertex (7 / 2, 100).

With y = 0, it gives:
x - 7 / 2 = +/- 5i
x = 7 / 2 +/- 5i.



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