Question

Find the real solutions, if any, of each equation. $$ \sqrt[4]{5 x^{2}-6}=x $$

Short answer

The real solutions are \ x = \pm \sqrt{2} \ and \ x = \pm \sqrt{3}.

Step-by-step solution

Understanding the Equation

We have the equation \[ \sqrt[4]{5x^2 - 6} = x \]. We are looking for values of x that satisfy this equation.

Eliminate the Fourth Root

To eliminate the fourth root, raise both sides of the equation to the fourth power: \[ \big(\sqrt[4]{5x^2 - 6}\big)^4 = x^4 \]. This simplifies to: \[ 5x^2 - 6 = x^4 \].

Rearrange the Equation

Move all terms to one side of the equation to set it equal to zero: \[ x^4 - 5x^2 + 6 = 0 \].

Substitute to Simplify

Let \[ y = x^2 \]. Then the equation becomes a quadratic in y: \[ y^2 - 5y + 6 = 0 \].

Solve the Quadratic Equation

Solve the quadratic equation \[ y^2 - 5y + 6 = 0 \] using factoring: \[ (y - 2)(y - 3) = 0 \]. The solutions are \[ y = 2 \] and \[ y = 3 \].

Back-Substitute x

Since \[ y = x^2 \], we have \[ x^2 = 2 \] and \[ x^2 = 3 \]. Solving for x, we get \[ x = \pm \sqrt{2} \] and \[ x = \pm \sqrt{3} \].

Verify Solutions

Verify that all four solutions satisfy the original equation. Plugging them back into the original equation confirms that \[ x = \sqrt{2}, \ - \sqrt{2}, \sqrt{3}, \ - \sqrt{3} \] are indeed solutions.

Key concepts

fourth roots

When you encounter a fourth root in an equation, it's useful to understand what this mathematical operation means.
The fourth root of a number is a value that, when raised to the power of four, yields the original number.
For example, the fourth root of 16 is 2 because 2 to the power of 4 is 16.
In the given equation \(\frac{1}{4}\big(5 x^2 - 6)=x\), the fourth root involves a little more complexity since it's nested within other algebraic terms.
It is also important to remember that raising both sides of an equation to an even power, such as four, can introduce extraneous solutions, so you must verify your solutions at the end.

solving quadratic equations

After simplifying the initial equation, we end up with \(5x^2 - 6 = x^4\), which, after some rearrangement, looks like \(x^4 - 5x^2 + 6 = 0\).
This is a higher-degree polynomial, but we can lower its degree by substituting \(y = x^2\).
The transformed equation \(y^2 - 5y + 6 = 0\) is a quadratic equation that we can solve using factoring, the quadratic formula, or completing the square.
Here, it's easier to factor: \((y - 2)(y - 3) = 0\).
Solving this, we find \(y = 2\) and \(y = 3\).
To go back to terms involving \(x\), remember that \(y = x^2\), so we find \(x^2 = 2\) and \(x^2 = 3\).

substitution method

The substitution method is a very useful algebraic strategy, especially when dealing with non-linear equations.
In the context of our problem, the substitution \(y = x^2\) turns a complicated polynomial equation into a simpler quadratic equation.
This simplification works wonders because quadratic equations are typically easier to solve.
Once you solve for \(y\), which represents \(x^2\), you can easily back-substitute to find values for \(x\).
It is crucial to remember to substitute back and check each solution within the context of the original equation, ensuring no extraneous solutions have been introduced.