Problem 64 - Algebra for College Students

Question

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

Step-by-Step Solution

Answer

Solutions are \[x = \frac{-1 \pm \sqrt{5}}{2}\] and \[x = \frac{-1 \pm \sqrt{17}}{2}\]

Step 1: Simplify the equation

Start by substituting a new variable to make the equation simpler. Set \[y = x^2 + x\]. Now the equation becomes \[y + \sqrt{y} - 2 = 0\]

Step 2: Isolate the square root term

Rewrite the equation to isolate the square root term: \[\sqrt{y} = 2 - y\]

Step 3: Square both sides

Square both sides of the equation to eliminate the square root: \[y = (2 - y)^2\]. This becomes: \[y = 4 - 4y + y^2\]

Step 4: Solve the quadratic equation

Re-arrange to obtain a quadratic equation: \[y^2 - 5y + 4 = 0\]. Factor the quadratic: \[(y - 1)(y - 4) = 0\]. This gives the solutions \[y = 1\] and \[y = 4\]

Step 5: Substitute back x

Substitute back \[y = x^2 + x\]. For \[y = 1\]: \[x^2 + x = 1\] solves to \[x^2 + x - 1 = 0\]. For \[y = 4\]: \[x^2 + x = 4\] solves to \[x^2 + x - 4 = 0\]

Step 6: Solve the quadratic equations for x

Solve the equations by the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. For \[x^2 + x - 1= 0\]: \[a = 1, b = 1, c = -1\]. Roots are: \[x = \frac{-1 \pm \sqrt{5}}{2}\]. For \[x^2 + x - 4 = 0\]: \[a = 1, b = 1, c = -4\]. Roots are: \[x = \frac{-1 \pm \sqrt{17}}{2}\]

Step 7: Combine all solutions

Combine all the solutions obtained from the two equations. Thus, the solutions are: \[x = \frac{-1 \pm \sqrt{5}}{2}\] and \[x = \frac{-1 \pm \sqrt{17}}{2}\]

Key Concepts

Quadratic Substitution

To simplify complex equations, sometimes we use a process called substitution. Here, we replace a complicated expression with a single variable. For example, the equation $$x^2 + x + \sqrt{x^2 + x} - 2 = 0$$ is tricky because of the square root. So, we let \[ y = x^2 + x \] This makes the equation easier: \[ y + \sqrt{y} - 2 = 0 \] Substitution helps break down the problem into simpler parts, making it more manageable.

Isolating Variables

After substitution, you often need to isolate one part of the equation. Here, we need to isolate the square root term \sqrt{y}. We do this by getting it by itself on one side of the equation: \[ y + \sqrt{y} - 2 = 0 \] becomes \[ \sqrt{y} = 2 - y \] Isolating variables helps us simplify further operations, such as squaring to eliminate the square root.

Quadratic Formula

The quadratic formula is a powerful tool to solve equations of the form \[ ax^2 + bx + c = 0 \] The formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For example, after isolating and squaring the variable, we solve \[ y^2 - 5y + 4 = 0 \] This is a quadratic equation. Factorizing gives: \[ (y - 1)(y - 4) = 0 \] Hence, \[ y = 1 \] or \[ y = 4 \] Next, we substitute back to solve for the original variable x using the quadratic formula. For instance: \[ x^2 + x - 1 = 0 \] becomes: \[ x = \frac{-1 \pm \sqrt{5}}{2} \] \[ x^2 + x - 4 = 0 \] becomes: \[ x = \frac{-1 \pm \sqrt{17}}{2} \]

Elimination of Square Roots

Eliminating square roots is another key step. It is often done by squaring both sides of the equation. For instance, when we have: \[ \sqrt{y} = 2 - y \] Squaring both sides gives: \[ (\sqrt{y})^2 = (2 - y)^2 \] Which simplifies to: \[ y = 4 - 4y + y^2 \] This allows us to transform the equation into one without square roots, making it easier to solve.