Problem 70 - Algebra and Trigonometry

Question

\(61-70\) . Find all solutions, real and complex, of the equation. $$ 1-\sqrt{x^{2}+7}=6-x^{2} $$

Step-by-Step Solution

Answer

Real solutions are \( x = 3 \) and \( x = -3 \).

Step 1: Setup the Equation

Start by isolating the square root on one side of the equation. The equation given is \[1 - \sqrt{x^2 + 7} = 6 - x^2.\] Subtract 1 from both sides to isolate the square root term:\[-\sqrt{x^2 + 7} = 5 - x^2.\]Next, multiply through by -1 to make the square root positive:\[ \sqrt{x^2 + 7} = x^2 - 5.\]

Step 2: Square Both Sides

To eliminate the square root, square both sides of the equation:\[ (\sqrt{x^2 + 7})^2 = (x^2 - 5)^2.\]This simplifies to:\[ x^2 + 7 = x^4 - 10x^2 + 25.\]

Step 3: Organize the Equation

Rearrange the terms to form a polynomial equation:\[ x^4 - 10x^2 - x^2 + 25 - 7 = 0.\]Simplify this to:\[ x^4 - 11x^2 + 18 = 0.\]

Step 4: Substitute and Solve

Let \( y = x^2 \), transforming the equation into a quadratic form:\[ y^2 - 11y + 18 = 0.\]Solve this quadratic equation using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -11 \), and \( c = 18 \).

Step 5: Apply the Quadratic Formula

The roots \( y \) are given by:\[ y = \frac{11 \pm \sqrt{11^2 - 4 \cdot 1 \cdot 18}}{2 \cdot 1}.\]Calculate the discriminant: \( 11^2 - 72 = 121 - 72 = 49 \).Thus, \[ y = \frac{11 \pm 7}{2}.\]This gives solutions \( y = 9 \) and \( y = 2 \).

Step 6: Back-Substitute for x

Recall that \( y = x^2 \). Therefore, solve for \( x \):1. \( x^2 = 9 \) gives \( x = 3 \) or \( x = -3 \).2. \( x^2 = 2 \) gives \( x = \sqrt{2} \) or \( x = -\sqrt{2} \).

Step 7: Verify the Solutions

Check each solution in the original equation:1. \( x = 3 \), substitute to check: \( 1-\sqrt{16} = 6-9 \Rightarrow -3 = -3 \) (Valid).2. \( x = -3 \), the same check applies.3. \( x = \sqrt{2} \), substitute: \( 1 - 3 = 6 - 2 \Rightarrow -2 eq 4 \) (Invalid).4. \( x = -\sqrt{2} \), same as above.Thus, only \( x = 3 \) and \( x = -3 \) are valid.

Key Concepts

Quadratic Formula

The quadratic formula is a reliable method for solving polynomial equations, specifically quadratic equations that take the form \( ax^2 + bx + c = 0 \). This method is extremely useful for equations where factoring is complex or impractical. By using the quadratic formula:

• Identify the coefficients \( a \), \( b \), and \( c \) from the quadratic equation.
• Calculate the discriminant using \( b^2 - 4ac \). The discriminant determines the nature of the roots, whether they are real or complex.
• Apply the formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find the solutions for \( x \).

In the context of our exercise, we transformed a higher degree equation into a quadratic form by substituting \( y = x^2 \). Then, utilizing the quadratic formula allowed us to find the values of \( y \), which were 9 and 2. This step simplifies solving equations involving squares or powers higher than two, making the quadratic formula indispensable in mathematics.

Polynomial Equations

Polynomial equations form the foundation of algebraic study, where equations consist of terms constructed from variables and coefficients with non-negative integer powers. These include linear, quadratic, cubic, and higher-order equations. The equation in this exercise, \( x^4 - 11x^2 + 18 = 0 \), is a quartic polynomial equation due to the highest power being four.

• First, understand that to solve polynomial equations, one of the strategies is transforming or simplifying it into valuable forms, like transforming it into a lower-degree polynomial.
• The substitution \( y = x^2 \) simplified our polynomial into a quadratic form, an easier equation to handle using known methods.
• This simplification step reduced the complexity significantly, making the equation manageable with the quadratic strategy for solving the roots.

Polynomial equations often require strategic manipulation to break down higher powers into solvable parts using substitution, factoring, or using formulas like the quadratic formula. This approach solidifies understanding and opens possibilities for solving more complex polynomial problems.

Solution Verification

Once you find potential solutions to an equation, it is crucial to verify these solutions by substituting them back into the original equation. This ensures they satisfy all conditions.

• Substitute each solution into the original equation to see if the left-hand side equals the right-hand side.
• Eliminate any extraneous solutions that do not satisfy the equation, as obtaining them is possible during manipulation like squaring both sides.
• For example, for the solutions \( x = 3 \) and \( x = -3 \), substituting them back confirmed they both satisfy the original equation \( 1 - \sqrt{x^2+7} = 6 - x^2 \).
• The solutions \( x = \sqrt{2} \) and \( x = -\sqrt{2} \) were found to be invalid upon verification.

This verification step is essential in problem-solving to avoid incorrect answers resulting from algebraic transformations, thereby ensuring the accuracy and validity of the solution.