### About Solving Equations that are Quadratic in Form:

We will encounter non-quadratic equations that are quadratic in form. We can create a quadratic equation by making a simple substitution. We can then solve the quadratic equation using the quadratic formula. When done, we substitute once more to obtain a solution in terms of the original variable involved.

Test Objectives

- Demonstrate the ability to identify a non-quadratic equation which is quadratic in form
- Demonstrate the ability to use substitution to create a quadratic equation
- Demonstrate the ability to solve an equation which is quadratic in form

#1:

Instructions: Solve each using the quadratic formula.

a) $$12x^4 + 7x^2 - 45 = 0$$

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#2:

Instructions: Solve each using the quadratic formula.

a) $$25x^4 - 10x^2 - 3 = 0$$

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#3:

Instructions: Solve each using the quadratic formula.

a) $$3x^\frac{2}{5} - 7x^\frac{1}{5} + 2 = 0$$

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#4:

Instructions: Solve each using the quadratic formula.

a) $$(7x^2 - 51)^2 - 8(7x^2 - 51) - 48 = 0$$

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#5:

Instructions: Solve each using the quadratic formula.

a) $$5x^\frac{2}{3} + 7x^\frac{1}{3} - 6 = 0$$

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Written Solutions:

#1:

Solutions:

a) $$x = \pm \frac{\sqrt{15}}{3} \hspace{.5em} or \hspace{.5em} x = \pm\frac{3i}{2}$$

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#2:

Solutions:

a) $$x = \pm\frac{\sqrt{15}}{5} \hspace{.5em} or \hspace{.5em} x = \pm\frac{i\sqrt{5}}{5}$$

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#3:

Solutions:

a) $$x = 32 \hspace{.5em} or \hspace{.5em} x = \frac{1}{243}$$

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#4:

Solutions:

a) $$x = \pm3 \hspace{.5em} or \hspace{.5em} x = \pm\frac{\sqrt{329}}{7}$$

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#5:

Solutions:

a) $$x = -8 \hspace{.5em} or \hspace{.5em} x = \frac{27}{125}$$