Problem 50 - Elementary Algebra

Question

Solve using the Square Root Property. \(2 d^{2}-4=77\)

Step-by-Step Solution

Answer

\(d = \pm \sqrt{40.5} \approx \pm 6.36\)

Step 1: Isolate the quadratic term

Add 4 to both sides of the equation to isolate the term with the variable: \[2d^2 - 4 + 4 = 77 + 4\] \[2d^2 = 81\]

Step 2: Divide by the coefficient of the quadratic term

Divide both sides by 2 to solve for \(d^2\): \[\frac{2d^2}{2} = \frac{81}{2}\] \[d^2 = \frac{81}{2}\]

Step 3: Apply the Square Root Property

Take the square root of both sides of the equation: \[d = \pm \sqrt{\frac{81}{2}}\] \[d = \pm \sqrt{40.5}\] \[d = \pm \sqrt{40.5}\]

Step 4: Simplify the square root

Simplify the square root to find the final solution: \[d = \pm \sqrt{40.5} \approx \pm 6.36\]

Key Concepts

Quadratic Equations

Quadratic equations are mathematical expressions where the highest power of the variable is 2. They usually take the form \(ax^2 + bx + c = 0\). The equation we have here, \(2d^2 - 4 = 77\), can be rearranged into this standard form by moving all terms to one side, resulting in \(2d^2 - 81 = 0\). The goal is to find the value of \(d\) that makes this statement true.
Quadratic equations can have two distinct solutions, one solution, or no real solutions, which depends on the discriminant (\backsqrt{b^2 - 4ac}\back). Here, we're focusing on solving the equation using the Square Root Property.

Solving Equations

Solving equations involves finding the value of the variable that makes the equation true. Let's break down the process step by step.
First, we need to isolate the term with the variable. In our case, adding 4 to both sides helps isolate the quadratic term: \(2d^2 - 4 + 4 = 77 + 4\) leads to \(2d^2 = 81\). Next, we need to simplify by removing the coefficient from the variable term by division: \( \frac{2d^2}{2} = \frac{81}{2} = 40.5\) simplifies to \(d^2 = 40.5\). Now, we apply the Square Root Property, which means taking the square root of both sides: \(d = \pm \backsqrt{40.5} \). This gives us the value of \(d \). Using square roots in this manner involves understanding how to handle both the positive and negative values that result from the square root operation.

Simplifying Square Roots

Simplifying square roots is the process of finding a more exact or simplified form of the number under the square root sign. Here, we simplified \( \sqrt{40.5}\) as accurately as possible.
When dealing with square roots, you often end up with irrational numbers. In our case, \( \sqrt{40.5} \) approximately equals 6.36. This is an approximate solution because 6.36 is not an exact number and represents an approximation.

• First, we notice that 40.5 can be expressed as \( \sqrt{40.5} \)
• Approximating to the nearest hundredth gives us \( d \approx \pm 6.36 \)

Hence the solution is \(d \approx \pm 6.36\), accurate to two decimal places.
Remember, when simplifying square roots, always look for factors that are perfect squares and try to simplify them further if possible.