Problem 79
Question
Solve each equation. $$x^{4}+2 x^{2}-15=0$$
Step-by-Step Solution
Verified Answer
The solutions are \(x = \sqrt{3}\) and \(x = -\sqrt{3}\).
1Step 1: Recognize the form of the equation
Identify that the given equation is in the form of a quadratic equation but with terms of higher power. The given equation is \(x^{4} + 2x^{2} - 15 = 0\).
2Step 2: Substitution
Substitute \(u = x^2\). This transforms the equation into a standard quadratic equation: \(u^2 + 2u - 15 = 0\).
3Step 3: Solve the quadratic equation
Solve the quadratic equation \(u^2 + 2u - 15 = 0\) by factoring or using the quadratic formula. Factoring, we get: \((u + 5)(u - 3) = 0\). So the solutions for \(u\) are \(u = -5\) and \(u = 3\).
4Step 4: Back-substitution
Replace \(u\) with \(x^2\) in the solutions: \(x^2 = -5\) and \(x^2 = 3\). Since \(x^2 = -5\) has no real solution, we discard it. We then have \(x^2 = 3\).
5Step 5: Solve for x
Solve for \(x\): \(x = \pm \sqrt{3} \). Therefore, the solutions to the original equation are \(x = \sqrt{3} \) and \( x = - \sqrt{3} \).
Key Concepts
Higher Order PolynomialsSubstitution MethodFactoring Quadratic EquationsSquare Roots
Higher Order Polynomials
Higher order polynomials are equations where the highest degree of the variable (like x) is greater than 2. In the given problem, the equation is
\[x^4 + 2x^2 - 15 = 0\].
This is a polynomial of degree 4 because the highest power of x is 4. To solve such equations without undue complexity, we often have to simplify them to familiar forms like quadratics.
In this case, substituting a variable (like u) for \( x^2 \) helps us reduce the higher order form to a more manageable quadratic equation.
\[x^4 + 2x^2 - 15 = 0\].
This is a polynomial of degree 4 because the highest power of x is 4. To solve such equations without undue complexity, we often have to simplify them to familiar forms like quadratics.
In this case, substituting a variable (like u) for \( x^2 \) helps us reduce the higher order form to a more manageable quadratic equation.
Substitution Method
The substitution method helps transform complex equations into simpler ones. For the given problem, we substituted \( u = x^2 \).
This transforms the original polynomial \(x^4 + 2x^2 - 15 = 0\) into a standard quadratic equation:
\[u^2 + 2u - 15 = 0\].
Using substitution, we converted a fourth-degree polynomial to a second-degree polynomial, making it easier to solve. Substitution simplifies higher-degree terms and lets us apply familiar solving methods.
This transforms the original polynomial \(x^4 + 2x^2 - 15 = 0\) into a standard quadratic equation:
\[u^2 + 2u - 15 = 0\].
Using substitution, we converted a fourth-degree polynomial to a second-degree polynomial, making it easier to solve. Substitution simplifies higher-degree terms and lets us apply familiar solving methods.
Factoring Quadratic Equations
Factoring quadratic equations is a method used to find the roots of the equation. The factored form of the equation makes it easy to solve for the variable.
For \(u^2 + 2u - 15 = 0\), we can factor this as: \((u + 5)(u - 3) = 0\).
This gives us two simple equations: \(u + 5 = 0\) and \(u - 3 = 0\).
From these, we find the solutions for u are \(u = -5\) and \(u = 3\). This step is crucial in reducing the quadratic to simpler equations where the solutions can be easily identified.
For \(u^2 + 2u - 15 = 0\), we can factor this as: \((u + 5)(u - 3) = 0\).
This gives us two simple equations: \(u + 5 = 0\) and \(u - 3 = 0\).
From these, we find the solutions for u are \(u = -5\) and \(u = 3\). This step is crucial in reducing the quadratic to simpler equations where the solutions can be easily identified.
Square Roots
Square roots help solve equations involving squared terms. Once we identify the solutions for u such as \(u = -5\) and \(u = 3\), we replace u back with \(x^2\). This gives us \(x^2 = -5\) and \(x^2 = 3\).
Since \(x^2 = -5\) has no real solutions, we discard it. For \(x^2 = 3\), taking the square root of both sides gives \(x = \pm \sqrt{3}\).
The solutions to the original equation are thus \(x = \sqrt{3}\) and \(x = -\sqrt{3}\). Square roots are essential in finding the final solutions after substituting back.
Since \(x^2 = -5\) has no real solutions, we discard it. For \(x^2 = 3\), taking the square root of both sides gives \(x = \pm \sqrt{3}\).
The solutions to the original equation are thus \(x = \sqrt{3}\) and \(x = -\sqrt{3}\). Square roots are essential in finding the final solutions after substituting back.
Other exercises in this chapter
Problem 79
Find each quotient. Write the answer in standard form \(a+b i .\) $$\frac{2}{3 i}$$
View solution Problem 79
Solve each rational inequality. Write each solution set in interval notation. $$\frac{7}{x+2} \geq \frac{1}{x+2}$$
View solution Problem 79
For each equation, ( \(a\) ) solve for \(x\) in terms of \(y,\) and ( \(b\) ) solve for \(y\) in terms of \(x\). $$4 x^{2}-2 x y+3 y^{2}=2$$
View solution Problem 80
Find each quotient. Write the answer in standard form \(a+b i .\) $$\frac{5}{9 i}$$
View solution