Question

Find the real solutions, if any, of each equation. $$ \sqrt[4]{4-3 x^{2}}=x $$

Short answer

The real solutions are \( x = 1 \) and \( x = -1 \).

Step-by-step solution

Understand the equation

The given equation is \( \sqrt[4]{4 - 3x^2} = x \). To solve for \(x\), isolate \(x\) and eliminate the radical by raising both sides to the fourth power.

Raise both sides to the fourth power

Raise both sides of the equation \( \left( \sqrt[4]{4 - 3x^2} \right)^4 = x^4 \) to eliminate the fourth root. The equation becomes \( 4 - 3x^2 = x^4 \).

Rearrange into a polynomial equation

Rearrange the equation to bring all terms to one side: \( x^4 + 3x^2 - 4 = 0 \). This is a polynomial equation in standard form.

Substitute to solve the polynomial

To solve \( x^4 + 3x^2 - 4 = 0 \), set \( y = x^2 \). The equation becomes \( y^2 + 3y - 4 = 0 \).

Factor the quadratic equation

Solve the quadratic equation \( y^2 + 3y - 4 = 0 \) by factoring: \( (y + 4)(y - 1) = 0 \).

Solve for \( y \)

Set each factor equal to zero: \( y + 4 = 0 \) or \( y - 1 = 0 \). Thus, \( y = -4 \) or \( y = 1 \).

Solve for \( x \) from each \( y \)

Since \( y = x^2 \), solve for \( x \) from the values of \( y \): 1. For \( y = -4 \): Since \( x^2 = -4 \) has no real solutions, discard this case. 2. For \( y = 1 \): Since \( x^2 = 1 \), \( x = \pm 1 \).

Verify the solutions

Substitute \( x = 1 \) and \( x = -1 \) back into the original equation to verify: 1. For \( x = 1 \), \( \sqrt[4]{4 - 3(1)^2} = 1 \) which is true. 2. For \( x = -1 \), \( \sqrt[4]{4 - 3(-1)^2} = -1 \) which is true.

Key concepts

Fourth Roots

The term 'fourth root' refers to the operation of \(\root 4\). If you have a number under the fourth root, it's the same as asking, 'What number multiplied by itself four times will result in this number?'. In the exercise, we see \(\root 4{4 - 3x^2} = x\). To eliminate the fourth root, raise both sides of the equation to the fourth power. This operation cancels the radical, simplifying the problem into a more familiar form. By transforming \( \root 4{A} \) into \((A)^4\), solving equations involving fourth roots can be more manageable.

Factoring Quadratic Equations

Factoring quadratic equations is a very useful skill in algebra. It's the process of breaking a quadratic down into the product of simpler binomial expressions. In the step-by-step solution, the equation \(y^2 + 3y - 4 = 0\) is factored into \((y + 4)(y - 1) = 0\). This turns the quadratic equation into two simpler equations: \(y + 4 = 0\) and \(y - 1 = 0\). When factored, it's much easier to find the roots or solutions of the equation. Factoring relies on finding two numbers that multiply to give the constant term and add to give the coefficient of the linear term.

Real Number Solutions

Real number solutions are the values of \(x\) that make the equation true within the set of real numbers. In the context of the exercise, when we solve \(y = x^2\), we focus on \(y = -4\) and \(y = 1\). Since \(x^2 = -4\) does not have a real number solution (the square of a real number cannot be negative), we discard it. \(x^2 = 1\) gives us two real solutions, \(\root 2{1} = 1\) and \(\root 2{-1} = -1\). Verifying these solutions confirm they satisfy the original equation.

Polynomial Equations

Polynomial equations are expressions that involve variables raised to whole number powers and coefficients. In the given solution, we transformed the original radical equation into a polynomial equation \((x^4 + 3x^2 - 4 = 0)\). Polynomial equations can often be solved by factoring, especially when they are in the quadratic form. Substituting \(y = x^2\) simplifies the polynomial to a more familiar quadratic equation \((y^2 + 3y - 4 = 0)\), which can be factored and solved more easily. This method allows for breaking down complex polynomial equations into simpler steps.