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Chapter 2: Problem 22

Find all real numbers \(x\) such that $$ x^{4}+5 x^{2}-14=0 $$.

Short Answer

Expert verified
The complete set of real solutions for \(x\) satisfying the equation \(x^4 + 5x^2 - 14 = 0\) are \(x_1 = \sqrt{2}\) and \(x_2 = -\sqrt{2}\).

Step by step solution

01

Replace \(x^2\) with \(y\)

Define a new variable \(y = x^2\). The equation then becomes \[y^2 + 5y - 14 = 0\]
02

Solve the quadratic equation

The given equation is now a quadratic equation. Solving for \(y\), we need to find the roots of the equation \[y^2 + 5y - 14 = (y - 2)(y + 7) = 0\] Thus, we have two possible values for \(y\): \[y_1 = 2 \quad \text{and} \quad y_2 = -7\]
03

Find the corresponding values of \(x\)

For each value of \(y\), we need to find the corresponding values of \(x\) such that \(x^2 = y\). For \(y_1 = 2\): \[x^2 = 2\] We find two real number solutions for \(x\): \[x_1 = \sqrt{2} \quad \text{and} \quad x_2 = -\sqrt{2}\] Now, for \(y_2 = -7\): \[x^2 = -7\] Since there is no real number \(x\) such that \(x^2 = -7\), we can conclude that there are no additional real solutions for \(x\) to the original equation.
04

List the complete set of real solutions

Finally, we have found the complete set of real solutions for \(x\) satisfying the equation \(x^4 + 5x^2 - 14 = 0\): \[x_1 = \sqrt{2}, \quad x_2 = -\sqrt{2}\]

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