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College Algebra : Equations Reducible to Quadratic Form

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Example Questions

Example Question #1 : Equations Reducible To Quadratic Form

Solve x^4-7x^2+12=0

Possible Answers:

$x=-\sqrt{3}$

x=-2

$x=\sqrt{3}$

x=2

$x=\pm\sqrt{3}$

$x=\pm2$

There are no solutions

Correct answer:

$x=\pm\sqrt{3}$

$x=\pm2$

Explanation:

To make this problem easier, lets start off by doing a u-substitution.

Let u=x^2 .

x^4-7x^2+12=0

u^2-7u+12=0

Now we can factor the left hand side.

(u-3)(u-4)=0

We have two solutions for , now we can plug those into u=x^2 , to get all the solutions.

$x^2=3\rightarrow x=\pm\sqrt{3}$

$x^2=4\rightarrow x=\pm\sqrt{4}=\pm2$

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Example Question #2 : Equations Reducible To Quadratic Form

Find all real roots of the polynomial function

f(x)=x^5-2x^3+x

Possible Answers:

$x = \left \{ 0,1\right \}$

$x = \left \{ -1,0,1\right \}$

$x = \left \{ -1,1\right \}$

x=0 (There are no nonzero solutions)

$x = \left \{ -2,-1,0,1,2\right \}$

Correct answer:

$x = \left \{ -1,0,1\right \}$

Explanation:

Find the roots of the polynomial,

f(x)=x^5-2x^3+x

Set f(x) equal to

x^5 -2x^3-x = 0

Factor out ,

x(x^4-2x^2-1)=0

Notice that the the factor x^4-2x^2-1 is a quadratic even though it might not seem so at first glance. One way to think of this is as follows:

Let u = x^2

Then we have u^2 = x^4 , substitute into x^4-2x^2-1 to get,

u^2-2u+1 = 0

Notice that the change in variable from to u=x^2 has resulted in a quadratic equation that can be easily factored due to the fact that it is a square of a simple binomial:

(u-1)^2=0

The solution for is,

u = 1

Because u = x^2 we go back to the variable ,

x^2 = 1

Therefore, the roots of the x^4-2x^2+1 factor are,

$x = \pm 1$

The other root of f(x) is x = 0 since the function clearly equals when x=0 .

The solution set is therefore, $x = \left \{ -1,0,1\right \}$

Below is a plot of f(x)=x^5 - 4x^3 +x . You can see where the function intersects the -axis at points corresponding to our solutions.

Further Discussion

The change of variable was a tool we used to write the quadratic factor in a more familiar form, but we could have just factored the original function in terms of as follows,

x(x^4-2x^2+1)

=x(x^2-1)^2

=x(x^2-1)(x^2-1)

=x(x-1)(x+1)(x-1)(x+1)

Setting this to zero gives the same solution set, $x = \left \{ -1,0,1\right \}$

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Example Question #2 : Equations Reducible To Quadratic Form

Give the complete solution set for the equation:

$x^{4} + 8x^{2} + 12 = 0$

Possible Answers:

$\left \{ -2 i , - i \sqrt{3}, i \sqrt{3} ,2 i \right \}$

$\left \{ - 2, - \sqrt{3}, \sqrt{3} ,2 \right \}$

$\left \{ - \sqrt{6}, - \sqrt{2}, \sqrt{2} , \sqrt{6} \right \}$

$\left \{ -2 i \sqrt{3}, - i , i ,2 i \sqrt{3}\right \}$

$\left \{ - i \sqrt{6}, - i \sqrt{2}, i \sqrt{2} , i \sqrt{6} \right \}$

Correct answer:

$\left \{ - i \sqrt{6}, - i \sqrt{2}, i \sqrt{2} , i \sqrt{6} \right \}$

Explanation:

$x^{4} + 8x^{2} + 12 = 0$

can be rewritten in quadratic form by setting $u = x^{2}$ , and, consequently, $u^{2} = x^{4}$ ; the resulting equation is as follows:

$u^{2} + 8u+ 12 = 0$

By the reverse-FOIL method we can factor the trinomial at left. We are looking for two integers with sum 8 and product 12; they are 2 and 6, so the equation becomes

(u+ 6)(u+ 2) = 0

Setting both binomials equal to 0, it follows that

u = -2 or u = -6 .

Substituting $x^{2}$ for , we get

$x^{2} = -2$ ,

in which case

$x = \pm i \sqrt{2}$ ,

or $x^{2} = -2$

in which case

$x = \pm i \sqrt{6}$ .

The solution set is $\left \{ - i \sqrt{6}, - i \sqrt{2}, i \sqrt{2} , i \sqrt{6} \right \}$ .

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Example Question #2 : Equations Reducible To Quadratic Form

Give the complete set of real solutions for the equation:

$e^{2x}- 8e^{x}+ 12 = 0$

Possible Answers:

$\left \{ \ln 3 , \ln 4 \right \}$

The equation has no solutions.

$\left \{ \ln 2, \ln 6 \right \}$

$\left \{ 0, \ln 12 \right \}$

Correct answer:

$\left \{ \ln 2, \ln 6 \right \}$

Explanation:

$e^{2x}- 8e^{x}+ 12 = 0$

can be rewritten in quadratic form by setting $u = e^{x}$ , and, consequently, $u^{2} =( e^{x})^{2} = e^{2x}$ ; the resulting equation is as follows:

$u^{2} - 8u+ 12 = 0$

By the reverse-FOIL method we can factor the trinomial at left. We are looking for two integers with sum 8 and product 12; they are and , so the equation becomes

(u-6)(u- 2) = 0

Setting both binomials equal to 0, it follows that

u =2 or u = 6 .

Substituting $e^{x}$ for , we get

$e^{x} = 2$

in which case

$x = \ln 2$ ,

and

$e^{x} = 6$ ,

in which case

$x = \ln 6$

The set of real solutions is therefore $\left \{ \ln 2, \ln 6 \right \}$ .

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College Algebra : Equations Reducible to Quadratic Form

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #1 : Equations Reducible To Quadratic Form

Example Question #2 : Equations Reducible To Quadratic Form

Example Question #2 : Equations Reducible To Quadratic Form

Example Question #2 : Equations Reducible To Quadratic Form

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