Problem 59 - College Algebra and Calculus: An Applied Approach

Question

Use a calculator to find the real solutions of the equation. (Round your answers to three decimal places.) \(3.2 x^{4}-1.5 x^{2}-2.1=0\)

Step-by-Step Solution

Answer

The real solutions for the equation are \(x = 0.586, x = -0.586\)

Step 1: Transformation into standard quadratic form

The given equation can be transformed into the form \(ay^2 + by + c = 0\) by letting \(x^2 = y\), yielding the new equation \(3.2y^2 - 1.5y - 2.1 = 0\).

Step 2: Application of quadratic formula

Use the quadratic formula to find the roots of \(y\).The root is \(y = [-(-1.5) ± sqrt((-1.5)^2 - 4*3.2*(-2.1))] / (2*3.2)\). After performing the calculation, we find that \(y = 0.343, -1.936\).

Step 3: Back substitution to find x

Since \(x^2 = y\), take the square root of both sides to find \(x\). The solutions are \(x = sqrt(0.343)\) and \(x = sqrt(-1.936)\). However, since taking square root of a negative number doesn't yield a real number, only \(sqrt(0.343)\) provides real solutions.

Step 4: Rounding results

After the square root calculation, the real solutions for \(x\) should be rounded to three decimal places as per the exercise instructions.

Key Concepts

Quadratic Formula

The quadratic formula is a key tool in finding solutions to quadratic equations, which are in the form \( ax^2 + bx + c = 0 \). This formula helps us solve for \(x\) by providing a straightforward method to determine the roots of the equation. The formula is expressed as:

• \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \)

Here, \(a\), \(b\), and \(c\) are coefficients in the quadratic equation. The plus-minus sign (\(\pm\)) indicates that there could be two potential solutions for \(x\). Using this formula, you can find both solutions quickly, even if the equation is complicated. The importance of the quadratic formula lies in its ability to solve any quadratic equation where the coefficients are real numbers, making it versatile for different problems.

Real Solutions

Real solutions in polynomial equations are values of \(x\) that satisfy the equation and are also real numbers. These solutions are crucial for understanding the behavior of quadratic equations because they determine if the graph of the equation will intersect the x-axis.

• A quadratic equation can have two real solutions, one real solution, or no real solutions if all solutions are complex numbers.
• To find real solutions, we solve the quadratic equation using the quadratic formula and check the discriminant \(b^2 - 4ac\).

If the discriminant is positive, there are two real solutions. If it is zero, there is one real solution. If the discriminant is negative, there are no real solutions because the roots will be complex numbers. In our problem, after transforming and solving \(3.2y^2 - 1.5y - 2.1 = 0\), the real solution was found when \(y = 0.343\), providing \(x = \sqrt{0.343}\).

Algebraic Transformation

Algebraic transformation involves changing an equation into a different format to make it easier to solve. In the given exercise, we transformed a quartic equation (an equation involving \(x^4\)) into a quadratic one using a substitution method.

• By setting \(x^2 = y\), the original equation \(3.2x^4 - 1.5x^2 - 2.1 = 0\) was simplified to \(3.2y^2 - 1.5y - 2.1 = 0\).

This transformation is crucial because solving quadratic equations is generally simpler and more straightforward than solving quartic equations. The transformation allows the application of the quadratic formula, thereby revealing the possible roots. This step is a powerful example of how algebraic manipulation can convert complex problems into manageable ones, highlighting the flexibility and power of algebraic processes.