Problem 11 - Algebra 2

Question

Solve each equation. $$ x^{4}-50 x^{2}+49=0 $$

Step-by-Step Solution

Answer

The solutions are \(x = 7, -7, 1, -1\).

Step 1: Identify the Equation Type

Recognize that the given equation \(x^4 - 50x^2 + 49 = 0\) is a quadratic in form because it can be rewritten as \((x^2)^2 - 50(x^2) + 49 = 0\). This suggests a substitution method can be used.

Step 2: Substitute a Variable

Let \(y = x^2\). Substitute \(y\) into the equation to transform it into a quadratic equation: \(y^2 - 50y + 49 = 0\).

Step 3: Solve the Quadratic Equation

Use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1, b = -50, c = 49\). Calculate \(b^2 - 4ac = (-50)^2 - 4 \cdot 1 \cdot 49 = 2500 - 196 = 2304\).

Step 4: Find the Roots

Calculate the roots using the quadratic formula: \(y = \frac{50 \pm \sqrt{2304}}{2}\). Since \(\sqrt{2304} = 48\), the roots are \(y = \frac{50 + 48}{2} = 49\) and \(y = \frac{50 - 48}{2} = 1\).

Step 5: Translate Back to \(x\)

Since \(y = x^2\), solve \(x^2 = 49\) and \(x^2 = 1\). This gives \(x = \pm 7\) and \(x = \pm 1\).

Step 6: Conclusion

Combine all solutions to state that the complete solution set for the original equation is \(x = 7, x = -7, x = 1, x = -1\).

Key Concepts

Quadratic Form

A polynomial equation is said to be in quadratic form when it can be expressed in the structure resembling a quadratic equation, even if the exponents of the actual variable do not strictly follow the typical quadratic powers. For example, in the equation \(x^4 - 50x^2 + 49 = 0\), it can be seen as quadratic in form because it can be rephrased as \[((x^2)^2 - 50(x^2) + 49 = 0)\]Let's break down why this restructure suggests it's quadratic in nature:

• The part \((x^2)^2\) indicates a square of a variable, which mirrors the \(x^2\) term seen in typical quadratics.
• The term \(-50(x^2)\) is linear in terms of \(x^2\), similar to the \(bx\) term in a quadratic equation (\(ax^2 + bx + c = 0\)).

The key to solving such equations is recognizing this structure, setting us up perfectly for the substitution method.

Substitution Method

The substitution method offers a clever way to simplify equations that initially seem complex. In the context of the equation \((x^2)^2 - 50(x^2) + 49 = 0\), we can simplify it by introducing a new variable. We set \(y = x^2\). Now our complicated equation becomes a more familiar quadratic equation: \[y^2 - 50y + 49 = 0\]Here's why substitution is helpful:

• This simplification transforms the original equation into a standard quadratic equation format, making it easier to solve using known methods.
• After solving for \(y\), we can easily revert back to terms of \(x\) because \(y\) was chosen such that \(y = x^2\).

This method is a powerful tool to tackle higher-degree polynomial equations by reducing them to quadratic equations.

Quadratic Formula

The quadratic formula is a universally applicable method used to find solutions to any quadratic equation of the form \(ax^2 + bx + c = 0\). The formula itself is expressed as:\[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]For our equation \(y^2 - 50y + 49 = 0\), we identify the coefficients:

• \(a = 1\)
• \(b = -50\)
• \(c = 49\)

Plugging these into the quadratic formula helps us find the values of \(y\). The term \(b^2 - 4ac\) under the square root (known as the discriminant) helps determine the nature of the roots:

• If it’s positive, the equation has two distinct real roots.
• If zero, it has exactly one real root.
• If negative, the roots are complex.

In our example, the discriminant \(b^2 - 4ac = 2304\) is positive, indicating two distinct real roots, which are calculated effectively using the quadratic formula.

Roots of Equation

Once we have solved for \(y\), we must determine the actual roots of the original polynomial equation in terms of \(x\). The roots derived from the quadratic equation \(y^2 - 50y + 49 = 0\) were found to be \(y = 49\) and \(y = 1\).This means:

• For \(y = 49\), since \(y = x^2\), solving \(x^2 = 49\) results in two solutions: \(x = 7\) and \(x = -7\).
• For \(y = 1\), solving \(x^2 = 1\) provides \(x = 1\) and \(x = -1\).

The complete solution set for the original polynomial equation is thus: \(x = 7, x = -7, x = 1, x = -1\).This step is crucial because it ties back the whole process to the initial equation, ensuring the transformations used didn’t lose any solutions along the way.