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Chapter 3: Problem 46

Find all real numbers \(x\) that satisfy the indicated equation. $$ x^{4}-8 x^{2}=-15 $$

Short Answer

Expert verified
The real numbers \(x\) that satisfy the given equation \(x^{4} - 8 x^{2} = -15\) are \(x = \pm \sqrt{3}, \pm \sqrt{5}\).

Step by step solution

01

Make a substitution

Let's substitute \(y\) for \(x^2\). We will have the equation: $$ y^2 - 8y = -15 $$
02

Rearrange the equation into standard quadratic form

Now, rearrange the equation into the standard quadratic form by adding 15 to both sides of the equation: $$ y^2 -8y + 15 = 0 $$
03

Factorize the quadratic equation

Next, factorize the quadratic equation to find the values of \(y\). This equation can be factored as: $$ (y - 3)(y - 5) = 0 $$
04

Solve for y

Now, we solve for \(y\). There are two possible solutions: $$ y - 3 = 0 \Rightarrow y = 3 $$ and $$ y - 5 = 0 \Rightarrow y = 5 $$
05

Re-substitute the value of y

Now remember that we made a substitution so that \(y = x^2\). Now we have two values for \(y\), so we need to solve for \(x\) by using each value of \(y\). For \(y = 3\): $$ x^2 = 3 \Rightarrow x = \pm \sqrt{3} $$ For \(y = 5\): $$ x^2 = 5 \Rightarrow x = \pm \sqrt{5} $$
06

Final solution is the union of the individual solutions

So, the final solution for the given equation is the union of the individual solutions we found when we solved for \(x\). Therefore, the values of \(x\) that satisfy the given equation are: $$ x = \pm \sqrt{3}, \pm \sqrt{5} $$

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