Problem 51 - Beginning and Intermediate Algebra

Question

Solve each equation. Check the solutions. \(x^{4}-29 x^{2}+100=0\)

Step-by-Step Solution

Answer

The solutions are \(x = 5, x = -5, x = 2, x = -2\).

Step 1: Substitute a New Variable

Let’s simplify the equation by substituting a new variable. Let’s set: \(y = x^{2}\)This transforms the original equation \(x^{4} - 29 x^{2} + 100 = 0\) into:\(y^{2} - 29 y + 100 = 0\)

Step 2: Solve the Quadratic Equation

We now solve the quadratic equation:\(y^{2} - 29 y + 100 = 0\)This can be solved using the quadratic formula: \[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, \(a = 1\), \(b = -29\), and \(c = 100\). Plugging in these values, we get:\[y = \frac{29 \pm \sqrt{841 - 400}}{2}\]\[y = \frac{29 \pm \sqrt{441}}{2}\]\[y = \frac{29 \pm 21}{2}\]Thus, \(y\) has two solutions:\[y = \frac{29 + 21}{2} = 25\]\[y = \frac{29 - 21}{2} = 4\]

Step 3: Substitute Back the Original Variable

Recall that \(y = x^{2}\). Now we have two equations to solve:1. \(x^{2} = 25\)2. \(x^{2} = 4\)Solving these gives us:\(x^{2} = 25\) \(\Rightarrow x = \pm 5\)\(x^{2} = 4\) \(\Rightarrow x = \pm 2\)

Step 4: Check All Solutions

We have four potential solutions: \(x = 5, x = -5, x = 2, x = -2\). We will substitute each back into the original equation to check:1. For \(x = 5\):\[ 5^{4} - 29 \cdot 5^{2} + 100 = 625 - 725 + 100 = 0\]2. For \(x = -5\):\[ (-5)^{4} - 29 \cdot (-5)^{2} + 100 = 625 - 725 + 100 = 0\]3. For \(x = 2\):\[ 2^{4} - 29 \cdot 2^{2} + 100 = 16 - 116 + 100 = 0\]4. For \(x = -2\):\[ (-2)^{4} - 29 \cdot (-2)^{2} + 100 = 16 - 116 + 100 = 0\]All solutions satisfy the original equation.

Key Concepts

quadratic substitution

Quadratic substitution is a helpful technique for solving polynomial equations, especially when faced with higher-degree polynomials. The idea is to simplify the equation by substituting a new variable to make it easier to solve. For instance, in the equation \(x^{4}-29x^{2}+100=0\), we can let \(y = x^{2}\). This converts our original equation into a quadratic form: \(y^{2}-29y+100=0\). By simplifying the problem, we can use familiar methods to find the solution. Quadratic substitution helps turn complex problems into simpler, more manageable ones.

checking solutions

After solving an equation, it is crucial to check all potential solutions. Initially, we found four possible solutions: \(x = 5\), \(x = -5\), \(x = 2\), and \(x = -2\). To ensure these are correct, we substitute each back into the original equation: \(x^{4} - 29x^{2} + 100 = 0\).
For \(x = 5\): \(5^4 - 29 \times 5^2 + 100 = 625 - 725 + 100 = 0\)
Testing other values like \(-5, 2\), and \(-2\) will show all satisfy the original equation. Checking solutions is essential to verify accuracy and guarantee no mistakes were made in the solving process.

quadratic formula

The quadratic formula is a powerful tool for solving quadratic equations. It is expressed as:
\[y = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\]
For our substituted quadratic problem \(y^{2}-29y+100=0\), we identify coefficients: \(a=1\), \(b=-29\), and \(c=100\). Substituting these values into the quadratic formula, we get:
\[y = \frac{29 \pm \sqrt{841 - 400}}{2} = \frac{29 \pm 21}{2}\]
This gives us two solutions: \(y=25\) and \(y=4\). The quadratic formula is extremely useful, offering a direct way to find the roots of any quadratic equation.

polynomial roots

Finding the roots of a polynomial means determining where the polynomial equals zero. For the problem \(x^{4}-29x^{2}+100=0\), we found the roots by breaking it down into simpler parts using quadratic substitution and solving the quadratic equation \(y^{2}-29y+100=0\).
After substitution, we obtained roots \(y=25\) and \(y=4\). Substituting back, we solved for \(x\):
\(x^{2}=25 \Rightarrow x= \pm5 \)
\(x^{2}=4 \Rightarrow x= \pm2\)
Thus, our original polynomial has roots \(x=5, x=−5, x=2, \ and \ x=−2\). Understanding polynomial roots helps in visualizing where the polynomial crosses the x-axis.