Problem 80 - Precalculus Student Solutions Manual 5th

Question

Solve each equation. $$3 x^{4}+10 x^{2}-25=0$$

Step-by-Step Solution

Answer

The solutions are \(x = \pm \frac{\sqrt{15}}{3}\).

Step 1: Substitute the variable

Let us make the substitution to simplify the equation. Let’s set \(y = x^2\). Hence, the equation becomes a quadratic equation in terms of \(y\):\[3y^2 + 10y - 25 = 0\]

Step 2: Apply the quadratic formula

Use the quadratic formula to solve for \(y\):\[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] where \(a = 3\), \(b = 10\), and \(c = -25\). Substituting in the values, we get:\[y = \frac{-10 \pm \sqrt{10^2 - 4(3)(-25)}}{2(3)}\]

Step 3: Simplify the discriminant

Simplify under the square root:\[10^2 - 4(3)(-25) = 100 + 300 = 400\]. Therefore, the quadratic formula becomes:\[y = \frac{-10 \pm \sqrt{400}}{6}\]

Step 4: Solve for y

Since \(\sqrt{400} = 20\), we can write:\[y = \frac{-10 + 20}{6} = \frac{10}{6} = \frac{5}{3}\]\[y = \frac{-10 - 20}{6} = \frac{-30}{6} = -5\]. Thus, \(y = \frac{5}{3}\) or \(y = -5\)

Step 5: Substitute back for x

Recall that \(y = x^2\), so we have \(x^2 = \frac{5}{3}\) or \(x^2 = -5\). Since \(x^2 = -5\) has no real solutions, we only consider \(x^2 = \frac{5}{3}\). Hence, \(x = \pm \sqrt{\frac{5}{3}}\)

Step 6: Simplify the solution

Express the solution for \(x\) in simplest form:\(x = \pm \frac{\sqrt{15}}{3}\)

Key Concepts

Quartic Equations

A quartic equation is a type of polynomial equation of degree four. These equations take on the general form of: \[ ax^4 + bx^3 + cx^2 + dx + e = 0 \] Here, the highest power of the variable (in this case, $$x$$) is four. Quartic equations can be challenging to solve, but they often can be simplified by various methods. In our example, the quartic equation is: \[3 x^4 + 10 x^2 - 25 = 0 \] Notice that there is no $$x^3$$ or $$x$$ term. This allows us to simplify it using a substitution method, which we will discuss next.

Substitution Method

The substitution method is a powerful algebraic tool to simplify complex polynomial equations. It involves substituting a part of the original equation with a new variable, making the equation easier to solve. In our given quartic equation: \[3 x^4 + 10 x^2 - 25 = 0\] we substitute $$x^2$$ with a new variable $$y$$. This turns the quartic equation into a quadratic one: \[3y^2 + 10y - 25 = 0\] Now, we can solve this simpler quadratic equation using the quadratic formula. Remember to substitute back once we solve for $$y$$ to find the original variable $$x$$.

Quadratic Formula

The quadratic formula is a standard method to solve quadratic equations of the form: \[ ax^2 + bx + c = 0 \] The formula is given by: \[ y = \frac{-b \, \pm \, \sqrt{b^2 - 4ac}}{2a} \] In our example, the quadratic equation is: \[3y^2 + 10y - 25 = 0\] Plugging in the values of $$a = 3$$, $$b = 10$$, and $$c = -25$$ into the quadratic formula, we get: \[ y = \frac{-10 \, \pm \, \sqrt{100 + 300}}{6} \] Simplifying further, we obtain: \[ y = \frac{-10 \, \pm \, \sqrt{400}}{6} \] Since $$\sqrt{400} = 20$$, the solutions for $$y$$ are: \[ y = \frac{10}{6} = \frac{5}{3}, \quad y = \frac{-30}{6} = -5 \] This results in two values for $$y$$: $$\frac{5}{3}$$ and $$-5$$.

Real Number Solutions

After solving for $$y$$, we substitute back to find the original variable $$x$$. Recall, $$y = x^2$$, so we have: \[ x^2 = \frac{5}{3} \quad \text{or} \quad x^2 = -5 \] The equation $$x^2 = -5$$ has no real number solutions because no real number squared equals a negative number. Thus, we discard $$y = -5$$. For $$ x^2 = \frac{5}{3}$$, we take the square root of both sides: \[ x = \pm \sqrt{\frac{5}{3}} \] Simplifying, we obtain: \[ x = \pm \frac{\sqrt{15}}{3} \] Hence, the real number solutions to the original quartic equation are: \[ x = \pm \frac{\sqrt{15}}{3} \] These are the values of $$x$$ that satisfy the given quartic equation.