Question

Find the real solutions of each equation. $$ 6 x^{4}-5 x^{2}-1=0 $$

Short answer

The real solutions are \( x = 1 \) and \( x = -1 \).

Step-by-step solution

Substitute Using a New Variable

Let’s make a substitution to simplify the equation. Let’s set \( y = x^2 \). This transforms our equation into a quadratic form:\[ 6y^2 - 5y - 1 = 0 \]

Use the Quadratic Formula

Next, solve the quadratic equation using the quadratic formula: \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 6 \), \( b = -5 \), and \( c = -1 \).Substitute these values into the formula:\[ y = \frac{5 \pm \sqrt{(-5)^2 - 4(6)(-1)}}{2(6)} = \frac{5 \pm \sqrt{25 + 24}}{12} = \frac{5 \pm \sqrt{49}}{12} \]

Solve for y

Solve the equation for \( y \):\[ y = \frac{5 + 7}{12} = 1 \quad \text{or} \quad y = \frac{5 - 7}{12} = -\frac{1}{6} \]

Restore the Original Variable

Now, revert to the original variable \( x \) by solving the equations \( x^2 = y \):For \( y = 1 \), we have \( x^2 = 1 \), which gives us: \( x = \pm 1 \).For \( y = -\frac{1}{6} \), since square values cannot be negative, there are no real solutions.

Combine the Solutions

The real solutions for our original equation are obtained from the valid values: \( x = 1 \) and \( x = -1 \).

Key concepts

Algebraic Substitution

Algebraic substitution is a powerful tool in solving complex equations by transforming them into simpler ones. In this exercise, we start with a polynomial equation:\( 6 x^{4} - 5 x^{2} - 1 = 0 \).
By substituting \( y = x^2 \), we simplify the equation into a quadratic form: \( 6 y^2 - 5 y - 1 = 0 \).
This substitution makes it easier to solve as we can now use quadratic solving techniques.
Substitution is particularly useful when dealing with higher-degree polynomials or when an equation is too complex to solve directly.
Always remember to revert back to the original variable at the end of your solution to find all possible values for the original problem.

Quadratic Formula

The quadratic formula is a standard method for solving quadratic equations of the form \( ax^2 + bx + c = 0 \).
In our exercise, we use the values \( a = 6 \), \( b = -5 \), and \( c = -1 \):

• Identify a, b, and c in the quadratic equation.
• Substitute them into the formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
• Simplify the expression under the root (discriminant) and solve for y.

The quadratic formula helps us find the precise values for \( y \) (or \( x \)). It is reliable and works for any quadratic equation, whether its solutions are real or complex.

Polynomial Equations

Polynomial equations involve terms with varying powers of the variable. In the given problem, the highest power is four (a quartic equation): \( 6 x^4 - 5 x^2 - 1 = 0 \).
Solving polynomial equations often requires a combination of methods, including:

• Factoring
• Substitution
• Using formulas like the quadratic formula

The key is to reduce the complexity of the equation, as we did here by substituting \( y = x^2 \), effectively transforming the quartic equation into a simpler quadratic equation.

Real Solutions

Solving an equation means finding all possible solutions that satisfy it. These solutions can be real or complex.
In this exercise, we seek the real solutions of the polynomial equation.
After substitution and using the quadratic formula, we find the solutions for \( y \): \( y = 1 \) and \( y = -\frac{1}{6} \).
Since \( y = x^2 \), we revert to the original variable:

• For \( y = 1 \), \( x^2 = 1 \) gives us \( x = \pm 1 \).
• For \( y = -\frac{1}{6} \), there are no real solutions because the square of a real number cannot be negative.

Hence, the real solutions of our original equation are \( x = 1 \) and \( x = -1 \). Real solutions are crucial when interpreting the practical application of mathematical models.