Question

Solve each equation algebraically. What are the restrictions on the variables? a) \(\sqrt{5 x^{2}+11}=x+5\) b) \(x+3=\sqrt{2 x^{2}-7}\) c) \(\sqrt{13-4 x^{2}}=2-x\) d) \(x+\sqrt{-2 x^{2}+9}=3\)

Short answer

a) No solutions. Restrictions: x >= -5; b) x = 8, -2. Restrictions: 2x^2 - 7 >= 0; c) x = 0.8. Restrictions: 2 - x >= 0; d) x = 0, 2. Restrictions: -2x^2 + 9 >= 0.

Step-by-step solution

for equation a - Isolate the square root term

First, ensure the square root term is isolated. For equation a: \[ \sqrt{5x^2 + 11} = x + 5 \]. It is already isolated in this case.

for equation a - Square both sides of the equation

Square both sides to eliminate the square root: \[ (\sqrt{5x^2 + 11})^2 = (x + 5)^2 \], which simplifies to \[ 5x^2 + 11 = x^2 + 10x + 25 \].

for equation a - Solve the quadratic equation

Rearrange terms to form a standard quadratic equation: \[ 5x^2 + 11 = x^2 + 10x + 25 \], which results in \[ 4x^2 - 10x - 14 = 0 \].Divide by 2: \[ 2x^2 - 5x - 7 = 0 \]. Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 2\), \(b = -5\), \(c = -7\).

for equation a - Calculate the roots

Calculate discriminant: \[ (-5)^2 - 4(2)(-7) = 25 + 56 = 81 \].Find roots: \[ x = \frac{5 \pm \sqrt{81}}{4} \Rightarrow x = 3 \text{ or } x = -\frac{7}{2} \].

for equation a - Verify solutions

Check both potential solutions in the original equation: For \(x = 3\): \[ \sqrt{5(3)^2 + 11} = \sqrt{56} eq 8 \]. For \(x = -\frac{7}{2}\): \[ \sqrt{5(-\frac{7}{2})^2 + 11} = \sqrt{54.25} eq -\frac{5}{2} \]. Both solutions are extraneous.

for equation a - State the restrictions

There are no solutions for \( \sqrt{5x^2 + 11} = x + 5 \). The restriction is \( x \geq -5 \) because the expression under the square root must be non-negative.

for equation b - Isolate the square root term

For equation b: \( x + 3 = \sqrt{2x^2 - 7} \). The square root term is already isolated.

for equation b - Square both sides of the equation

Square both sides to eliminate the square root: \((x + 3)^2 = (\sqrt{2x^2 - 7})^2 \), which simplifies to \( x^2 + 6x + 9 = 2x^2 - 7 \).

for equation b - Solve the quadratic equation

Rearrange terms: \( x^2 + 6x + 9 = 2x^2 - 7 \) results in \( x^2 - 6x - 16 = 0 \). Use the quadratic formula with \(a = 1\), \(b = -6\), \(c = -16\).

for equation b - Calculate the roots

Calculate discriminant: \[(-6)^2 - 4(1)(-16) = 36 + 64 = 100\]. Find roots: \[ x = \frac{6 \pm \sqrt{100}}{2} \Rightarrow x = 8\text{ or }x = -2\].

for equation b - Verify solutions

Check both potential solutions in the original equation: For \(x = 8\): \( 8 + 3 = \sqrt{2(8)^2 - 7} \Rightarrow 11 = \sqrt{121} = 11 \). For \(x = -2\): \( -2 + 3 = \sqrt{2(-2)^2 - 7} \Rightarrow 1 = \sqrt{1} = 1 \). Both solutions are valid.

for equation b - State the restrictions

The solutions are \(x = 8 \text{ and } x = -2\). The restriction is \( 2x^2 - 7 \geq 0\).

for equation c - Isolate the square root term

For equation c: \(\sqrt{13 - 4x^2} = 2 - x \). The square root term is already isolated.

for equation c - Square both sides of the equation

Square both sides to eliminate the square root: \((\sqrt{13 - 4x^2})^2 = (2 - x)^2 \), which simplifies to \( 13 - 4x^2 = 4 - 4x + x^2 \).

for equation c - Solve the quadratic equation

Rearrange terms: \( 13 - 4x^2 = 4 - 4x + x^2 \) results in \( 5x^2 - 4x - 9 = 0 \). Use the quadratic formula with \(a = 5\), \(b = -4\), \(c = -9\).

for equation c - Calculate the roots

Calculate discriminant: \[(-4)^2 - 4(5)(-9) = 16 + 180 = 196\].Find roots: \[ x = \frac{4 \pm \sqrt{196}}{10} \Rightarrow x = \frac{8}{10} \text{ or } x = -\frac{-12}{10} \Rightarrow x = 0.8 \text{ or } x = -1.2 \].

for equation c - Verify solutions

Check both potential solutions in the original equation: For \(x = 0.8\): \[\sqrt{13 - 4(0.8)^2} = 2 - 0.8 \Rightarrow \sqrt{10.44} \approx 1.2 \].For \(x = -1.2\): \[\sqrt{13 - 4(-1.2)^2} = 2 + 1.2 \Rightarrow \sqrt{8.24} \approx 3.2 \]. Only \(x = 0.8\) is valid.

for equation c - State the restrictions

The solution is \(x = 0.8\). The restriction is \( 2 - x \geq 0 \).

for equation d - Isolate the square root term

For equation d: \(x + \sqrt{-2x^2 + 9} = 3 \). Isolate the square root by rearranging: \( \sqrt{-2x^2 + 9} = 3 - x \).

for equation d - Square both sides of the equation

Square both sides to eliminate the square root: \((3 - x)^2 = (-2x^2 + 9) \), which simplifies to \(9 - 6x + x^2 = -2x^2 + 9 \).

for equation d - Solve the quadratic equation

Rearrange terms: \( 9 - 6x + x^2 = -2x^2 + 9 \) results in \( 3x^2 - 6x = 0 \). Solve by factoring: \( 3x(x - 2) = 0\).

for equation d - Calculate the roots

Set each factor to zero: \(3x = 0 \Rightarrow x = 0\) and \(x - 2 = 0 \Rightarrow x = 2\).

for equation d - Verify solutions

Check both potential solutions in the original equation: For \(x = 0\): \[0 + \sqrt{-2(0)^2 + 9} = 3 \Rightarrow \sqrt{9} = 3 \]. For \(x = 2\): \[2 + \sqrt{-2(2)^2 + 9} = 3 \Rightarrow 2 + \sqrt{1} = 3 \Rightarrow 2 + 1 = 3 \]. Both solutions are valid.

for equation d - State the restrictions

The solutions are \(x = 0 \text{ and } x = 2\). The restriction is \( -2x^2 + 9 \geq 0 \).

Key concepts

quadratic equations

Quadratic equations are a type of polynomial equation where the highest power of the variable is 2. These equations can be written in the standard form \[ ax^2 + bx + c = 0 \], where \(a, b\), and \(c\) are constants, and \((a eq 0)\). To solve them, we often use methods like factoring, completing the square, or utilizing the quadratic formula. For example, the quadratic formula is \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \], which can solve any quadratic equation. Various techniques may need to be combined when solving complex or specially formulated quadratics, especially those involving square roots.

algebraic solutions

Algebraic solutions involve manipulating equations using algebraic operations to isolate the variable and solve for its value. To solve quadratic equations, the following steps are generally taken:

• Isolate terms: This often involves moving like terms to one side of the equation.


• Eliminate square roots: By squaring both sides of the equation, we can remove the square root operator, leading to a simpler polynomial equation.


• Solve quadratic formula: Standard methods such as factoring, completing the square, or using the quadratic formula are then employed.

Algebraic solutions ensure the proper handling of variables and constants to simplify the equations. These steps help break down the problem, making it easier to find the solution.

square roots

Square roots are the inverse operation to squaring a number. When solving equations like \(\sqrt{A} = B \), squaring both sides can help eliminate the square root. However, one must be cautious as this step can sometimes introduce extraneous solutions.

For instance, consider the equation \(\sqrt{5x^2 + 11} = x + 5\). By squaring both sides, we simplify to \(5x^2 + 11 = (x + 5)^2\) and further reduce it to solve for \(x\). Understanding and correctly applying the properties of square roots is crucial in getting accurate solutions.

extraneous solutions

Extraneous solutions are solutions that arise from the algebraic manipulation of an equation but do not satisfy the original equation. This frequently happens when both sides of an equation are squared or when the equation involves square roots.

For example, consider the equation \(\sqrt{5x^2 + 11} = x + 5\). After squaring both sides and solving, potential solutions are \(x = 3\) and \(x = -\frac{7}{2}\). However, when substituting these values back into the original equation, neither satisfies it, thus both are extraneous.
This emphasizes the importance of verifying solutions against the initial equation to ensure they are valid. Always check your solutions to avoid including any extraneous ones.