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Chapter 10: Problem 49

Solve using the Square Root Property. \(6 c^{2}+4=29\)

Short Answer

Expert verified
c = \pm \frac{5\sqrt{6}}{6}

Step by step solution

01

Isolate the Quadratic Term

Start by isolating the quadratic term on one side of the equation. Subtract 4 from both sides to get: \[ 6c^2 + 4 - 4 = 29 - 4 \]\[ 6c^2 = 25 \]
02

Divide by the Coefficient of the Quadratic Term

Divide both sides by the coefficient of the quadratic term, which is 6, to simplify the equation:\[ \frac{6c^2}{6} = \frac{25}{6} \]\[ c^2 = \frac{25}{6} \]
03

Apply Square Root Property

Apply the Square Root Property to both sides of the equation to solve for c:\[ c = \pm \sqrt{\frac{25}{6}} \]
04

Simplify the Square Root

Simplify the square root to get the final solution. We know that the square root of a fraction is the square root of the numerator divided by the square root of the denominator:\[ c = \pm \frac{\sqrt{25}}{\sqrt{6}} \]Since \(\sqrt{25} = 5\), the solution is:\[ c = \pm \frac{5}{\sqrt{6}} \]Rationalize the denominator to get:\[ c = \pm \frac{5\sqrt{6}}{6} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

solving quadratic equations
Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\). In this exercise, we have an equation \(6c^2 + 4 = 29\).

The goal is to isolate the squared term. This process involves:
  • Moving all other terms to one side.
  • Simplifying the equation to make it easier to solve.

For instance, subtract 4 on both sides to get \(6c^2 = 25\). After isolating the quadratic term, we can solve for \(c\).
square root property
The Square Root Property is a method used to solve quadratic equations that do not have a linear term. Once the quadratic term is isolated and simplified, we apply the Square Root Property: if \(x^2 = k\), then \(x = \pm\sqrt{k}\).

For our equation, after isolating and simplifying, we get \(c^2 = \frac{25}{6}\). Applying the Square Root Property gives us two possible solutions: \(c = \pm\sqrt{\frac{25}{6}}\).

Always remember to include both the positive and negative roots.
simplifying radicals
Simplifying radicals means writing them in their simplest form. For our equation, we have \(c = \pm\sqrt{\frac{25}{6}}\).

To simplify, we take the square root of both numerator and denominator: \( c = \pm\frac{\sqrt{25}}{\sqrt{6}}\). Since \(\sqrt{25} = 5\), the equation simplifies to \( c = \pm\frac{5}{\sqrt{6}}\).

Use this method to break down complex radicals, making them easier to work with.
rationalizing the denominator
Rationalizing the denominator is the process of removing radicals from the denominator of a fraction. Our simplified solution is \(c = \pm\frac{5}{\sqrt{6}}\).

To rationalize, we multiply both numerator and denominator by the radical in the denominator:
\(c = \pm\frac{5}{\sqrt{6}}\cdot\frac{\sqrt{6}}{\sqrt{6}} = \pm\frac{5\sqrt{6}}{6}\).

This process gives us a fraction with a rational number in the denominator, which is generally preferred in mathematics.

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Most popular questions from this chapter

Solve by using the Quadratic Formula. \(3 q^{2}+8 q-3=0\)

In the following exercises, determine the number of solutions to each quadratic equation. a. \(25 p^{2}+10 p+1=0\) b.\(7 q^{2}-3 q-6=0\) c.\(7 y^{2}+2 y+8=0\) d.\(25 z^{2}-60 z+36=0\)

In the following exercises, solve. Round answers to the nearest tenth. A rancher is going to fence three sides of a corral next to a river. He needs to maximize the corral area using 240 feet of fencing. The quadratic equation \(A=x(240-2 x)\) gives the area of the corral, \(A\), for the length, \(x,\) of the corral along the river. Find the length of the corral along the river that will give the maximum area, and then find the maximum area of the corral.

Solve by completing the square. \(2 p^{2}+7 p=14\)

In the following exercises, find the maximum or minimum value. $$ y=x^{2}-6 x+15 $$
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