1. Solving Equations that can be rewritten as
quadratic equations
• These are solved through the use of substitution
Ex. 2x4 ­ 5x2 + 2 = 0
Temporarily substitute p in place of x2
You are now solving a quadratic for p
In the end solve for x by
x2 = p
so therefore
x = ± √p
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2. Ex. 2x4 ­ 5x2 + 2 = 0
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3. Properties of the roots(solutions)
of a Quadratic Equation
The average of the sum of the roots gives you the value
of x at the vertex, and the axis of symmetry
Therefore, if we are given the roots, we can find the
vertex using
x at the vertex =
Therefore y at the vertex = f( )
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4. Ex. Find the vertex of the parabola given by
x2 ­ 6x ­7 =0
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5. You can also generate the equation of a quadratic if you are given
the roots
The equation can be found by
x2 ­ (sum of the roots)x + (product of the roots) = 0
Show that the sum of the roots =
and
Show that the product of the roots =
This is useful because our quadratic equation can be then written as
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6. Sum of the roots
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7. Product of the roots
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8. 8

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