Question

Find the real solutions, if any, of each equation. $$ \sqrt{2 x+5}-x=1 $$

Short answer

The solution is \( x = 2 \).

Step-by-step solution

Isolate the Square Root

Start by isolating the square root on one side of the equation. Add \( x \) to both sides: \[ \sqrt{2x + 5} = x + 1 \]

Square Both Sides

To eliminate the square root, square both sides of the equation: \[ ( \sqrt{2x + 5} )^2 = ( x + 1 )^2 \] which simplifies to \[ 2x + 5 = ( x + 1 )^2 \]

Expand the Right Side

Expand the right side of the equation: \[ 2x + 5 = x^2 + 2x + 1 \]

Set Up a Quadratic Equation

Rearrange the equation to set it to 0: \[ x^2 + 2x + 1 - 2x - 5 = 0 \] which simplifies to \[ x^2 - 4 = 0 \]

Solve the Quadratic Equation

Factor the quadratic equation: \[ (x - 2)(x + 2) = 0 \] Set each factor to zero and solve for \( x \): \[ x - 2 = 0 \] hence \( x = 2 \) \[ x + 2 = 0 \] hence \( x = -2 \)

Verify the Solutions

Substitute \( x = 2 \) into the original equation: \( \sqrt{2(2) + 5} - 2 = 1 \) \( \sqrt{9} - 2 = 1 \) \( 3 - 2 = 1 \) ✓ valid solution. Substitute \( x = -2 \) into the original equation: \( \sqrt{2(-2) + 5} - (-2) = 1 \) \( \sqrt{1} + 2 = 1 \) \( 1 + 2 ≠ 1 \) invalid solution.

Key concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation of the form \( ax^2 + bx + c = 0 \) where \( a, b, \) and \( c \) are constants and \( a eq 0 \).
A key property of a quadratic equation is that it graphs into a parabola, which can open upwards or downwards depending on the sign of \( a \). Here’s a step-by-step guide to understand quadratic equations:

• Identify the coefficients \( a, b, \) and \( c \) from the equation.
• The solutions, or roots, of the equation can be found using factoring, completing the square, or the quadratic formula \( x = \frac{-b \, \text{±} \, \text{√}(b^2 - 4ac)}{2a} \).

In our problem, during the steps of isolating the square root and squaring both sides, we end up with \( x^2 - 4 = 0 \). This represents a quadratic equation.
The roots can be determined either through factoring or using the quadratic formula.

Square Root

The square root operation is the inverse of squaring a number. Essentially, for a given non-negative number \( y \), the square root is a number \( x \) such that \( x^2 = y \).
When dealing with equations involving square roots, the first step is typically to isolate the square root on one side. This makes the process of solving easier by allowing you to square both sides to eliminate the square root.
In our exercise, we began by isolating the square root in the equation \( \text{√}(2x + 5) - x = 1 \). This was done by rearranging the terms to get \( \text{√}(2x + 5) = x + 1 \).
Next, squaring both sides to remove the square root converts the equation into a polynomial form that can be solved using other algebraic methods, leading to our quadratic equation.

Factoring

Factoring is one of the common methods used to solve quadratic equations. When you factor a quadratic equation, you’re expressing it as a product of two binomials.
The general form for factoring a quadratic equation \( ax^2 + bx + c \) is \( (dx + e)(fx + g) = 0 \).
Here are the steps to factor a quadratic equation:

• Write down the equation in standard form: \( ax^2 + bx + c \).
• Identify two numbers that multiply to give \( ac \) and add to give \( b \).
• Use these numbers to break up the middle term \( bx \) and factor by grouping.

In our exercise, the quadratic equation formed was \( x^2 - 4 = 0 \). The factors of \( -4 \) that add up to \( 0 \) are \( 2 \) and \( -2 \). So, we can factorize as \( (x - 2)(x + 2) = 0 \).
Setting each factor to zero provides the solutions to the equation: \( x = 2 \) and \( x = -2 \). Verification helps in determining which solutions are valid as we did in our last steps.