Problem 63 - Algebra for College Students

Question

Find all real solutions to each equation. $$2 x^{2}-3-6 \sqrt{2 x^{2}-3}+8=0$$

Step-by-Step Solution

Answer

The solutions are \( x = \pm \sqrt{\frac{19}{2}} \) and \( x = \pm \sqrt{\frac{7}{2}} \).

Step 1 - Simplify the equation

Start by simplifying the given equation: \[2x^2 - 3 - 6\sqrt{2x^2 - 3} + 8 = 0 \]Combine like terms: \[2x^2 + 5 - 6\sqrt{2x^2 - 3} = 0\]

Step 2 - Make a substitution

Let \( y = \sqrt{2x^2 - 3} \). Then, square both sides to find \( y^2 \): \[ y^2 = 2x^2 - 3 \]Substitute \( y \) into the original equation: \[ 2x^2 + 5 = 6y \]

Step 3 - Relate the two equations

From \( y^2 = 2x^2 - 3 \), express \( 2x^2 \) as: \[2x^2 = y^2 + 3 \]Substitute \( 2x^2 \) from this equation back into the modified original equation: \[ y^2 + 3 + 5 = 6y \]Thus, simplify the equation to get: \[ y^2 + 8 = 6y \]Rearrange to form a standard quadratic equation: \[ y^2 - 6y + 8 = 0 \]

Step 4 - Solve the quadratic equation

Solve the quadratic equation \( y^2 - 6y + 8 = 0 \) using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Plug in the constants \( a = 1 \), \( b = -6 \), and \( c = 8 \): \[ y = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1} \]Simplify under the square root: \[ y = \frac{6 \pm \sqrt{36 - 32}}{2} \]\[ y = \frac{6 \pm \sqrt{4}}{2} \]\[ y = \frac{6 \pm 2}{2} \]Thus, \( y = 4 \) or \( y = 2 \)

Step 5 - Back substitute to find \( x \)

Recall the substitution \( y = \sqrt{2x^2 - 3} \). Thus: For \( y = 4 \): \[ 4 = \sqrt{2x^2 - 3} \]Square both sides: \[ 16 = 2x^2 - 3 \]\[ 19 = 2x^2 \]\[ x^2 = \frac{19}{2} \]\[ x = \pm \sqrt{\frac{19}{2}} \]For \( y = 2 \): \[ 2 = \sqrt{2x^2 - 3} \]Square both sides: \[ 4 = 2x^2 - 3 \]\[ 7 = 2x^2 \]\[ x^2 = \frac{7}{2} \]\[ x = \pm \sqrt{\frac{7}{2}} \]

Key Concepts

quadratic substitution

Quadratic substitution is a powerful technique used to simplify complex equations. In this method, we replace a part of the original equation with a new variable, making it easier to solve. For instance, in our given problem, we let \( y = \sqrt{2x^2 - 3} \). This turns the original equation into a simpler form. By making this substitution, we reduce a complicated expression into a quadratic form that is easier to handle.
After substituting, we find two equations that relate \( y \) and \( x \). These simpler equations can then be solved using standard methods for quadratic equations. Once we find values for \( y \), we can back-substitute to find the original variable \( x \).

standard quadratic equation

A standard quadratic equation is any equation of the form \( ax^2 + bx + c = 0 \). In our problem, after substitution and simplification, we obtained the quadratic equation \( y^2 - 6y + 8 = 0 \). This is clearly in the standard form where \( a = 1 \), \( b = -6 \), and \( c = 8 \).

The standard form is crucial because it allows us to use a variety of solving techniques like factoring, completing the square, or using the quadratic formula. Understanding this form is the foundation for solving many types of algebraic problems.

quadratic formula

The quadratic formula is a universal method for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). The formula is given by:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula is derived from the process of completing the square. It allows us to find the roots of any quadratic equation, provided we know the coefficients \( a \), \( b \), and \( c \).
In our specific problem, we applied the quadratic formula to find the values of \( y \) with \( a = 1 \), \( b = -6 \), and \( c = 8 \). Substituting these values, we got:
\[ y = \frac{6 \pm \sqrt{36 - 32}}{2} \]
This simplifies to:
\[ y = \frac{6 \pm 2}{2} \]
resulting in solutions \( y = 4 \) and \( y = 2 \).

back substitution

Back substitution is the final step in the process, where we revert our earlier substitution to find the original variable's value(s). In our problem, we had made the substitution \( y = \sqrt{2x^2 - 3} \). Now, we use the values of \( y \) we obtained (\( y = 4 \) and \( y = 2 \)) to find \( x \).
For \( y = 4 \), we solve:
\[ 4 = \sqrt{2x^2 - 3} \]
Squaring both sides, we get:
\[ 16 = 2x^2 - 3 \]
Which simplifies to:
\[ 19 = 2x^2 \]
Therefore,
\[ x^2 = \frac{19}{2} \]
and
\[ x = \pm \sqrt{\frac{19}{2}} \]Similarly, for \( y = 2 \), the steps lead to:
\[ x = \pm \sqrt{\frac{7}{2}} \]

This final step translates the answers from the substituted form back into the original variable's terms.