Problem 54 - College Algebra Essentials

Question

By making an appropriate substitution. $$x+3 x^{\frac{1}{2}}-4=0$$

Step-by-Step Solution

Answer

The only solution of the given equation is \( x = 1 \).

Step 1: Identify the substitution

We note that \( x = (x^{\frac{1}{2}})^2 \). So we set \( u = x^{\frac{1}{2}} \), which simplifies the original equation to \( u^2 + 3u - 4 = 0 \).

Step 2: Solve the simplified equation

The equation can be factorized into \( (u-1)(u+4) = 0 \). Setting each factor equal to zero gives the solutions \(u=1\) and \(u=-4\).

Step 3: Revert the Substitution

To get the solutions in terms of x, replace \( u \) with \( x^{\frac{1}{2}} \). Which gives \( x^{\frac{1}{2}} = 1 \) and \( x^{\frac{1}{2}} = -4 \). Solving these gives \( x = 1^2 = 1 \) and the second equation has no real solution since \( x \) must be nonnegative.

Key Concepts

Understanding the Substitution Method

The substitution method is an essential technique used to simplify complex equations. It's particularly helpful when dealing with equations involving radicals, like square roots. In the original exercise, you're given an equation:
\[x + 3x^{\frac{1}{2}} - 4 = 0\]
To aid in understanding and solving this equation, a substitution is made. Notice that \({x = (x^{\frac{1}{2}})^2}\). By setting \(u = x^{\frac{1}{2}}\), the equation transforms into:

• \[u^2 + 3u - 4 = 0\]

This new equation is simpler because it removes the square root, turning it into a standard quadratic form. Substitution helps in reducing the complexity, making the task of finding the solutions more straightforward.
With substitution, ensure that you clearly define the variable you're substituting. In this example, marking \({u}\) as \({x^{\frac{1}{2}}}\) is crucial. It sets the foundation for solving the quadratic equation effectively.

Factoring Quadratic Equations

Factoring is a method used to solve quadratic equations, expressed in the form \(ax^2 + bx + c = 0\). The key is to express the quadratic as a product of two binomials. Let's observe how factoring applies to our simplified equation:

• \[u^2 + 3u - 4 = 0\]

By examining the factors of \(ac\) that yield \(b\), we discover the factors \(1\) and \(-4\). Therefore, the quadratic can be factored into:

• \((u-1)(u+4) = 0\)

When the equation is factored, each binomial is set to zero to solve for \(u\):

• \(u - 1 = 0\rightarrow u = 1\)
• \(u + 4 = 0\rightarrow u = -4\)

Factoring is powerful because it simplifies solving the equation by breaking it down into smaller parts. The solutions \(u = 1\) and \(u = -4\) are found by each factor equaling zero, making it straightforward to interpret.

Solving Equations Involving Square Roots

Solving equations with square roots involves additional steps because of the nature of the square root. After using substitution and factoring in our problem, we have the solutions for \(u\):

• \(u = 1\)
• \(u = -4\)

Recall that \(u\) was defined as \(x^{\frac{1}{2}}\). Let's now revert our substitution and solve for \(x\). For \(u = 1\):

• \(x^{\frac{1}{2}} = 1\)
• Squaring both sides, \(x = 1^2 = 1\)

However, when \(u = -4\), the equation \(x^{\frac{1}{2}} = -4\) does not have a real solution because a square root of a real number cannot be negative.
When solving such equations, remember:

• Check if the solutions fit the constraints of square roots (e.g., non-negativity).
• Return to the original variable after solving.

This process ensures that the roots you find are valid in the context of the original equation.