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Chapter 1: Problem 17

Solve each equation in Exercises \(15-26\) by the square root method. $$5 x^{2}+1=51$$

Short Answer

Expert verified
The solutions of the exercise are \( x = \sqrt{10} \) and \( x = -\sqrt{10} \)

Step by step solution

01

Isolate the squared term

First you isolate the square term \(5x^2\) on one side of the equation by subtracting 1 from both sides: \[5x^2 = 51 - 1\] This simplifies to \[5x^2 = 50\]
02

Remove the coefficient from the variable

To remove the coefficient of 5 from the variable, you have to divide both sides of the equation by 5: \[x^2 = \frac{50}{5}\] that becomes \[x^2 = 10\]
03

Solve for variable x using the square root

Now, use the square root method to solve for x. Remember that you must take into account both the positive and negative square root. We have, \[x = \pm \sqrt{10}\]

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Key Concepts

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

Square Root Method
The square root method is a handy technique for solving quadratic equations when the equation is in the form \[ ax^2 + c = 0. \] This approach involves taking the square root of both sides of the equation to find the value of \( x \). The sweetness of this method lies in its simplicity: it can sidestep the more complex operations usually required for quadratics. But remember, every square has two roots鈥攁 positive and a negative. Hence, always include + \( \sqrt{} \) and - \( \sqrt{} \) in your result.For instance, let's consider: \[ x^2 = 10. \]The square root of \( 10 \) yields approximately \( 3.16 \). Hence, \( x \) can be either \( +3.16 \) or \( -3.16 \).
  • Positive Square Root: \( x = +\sqrt{10} \)
  • Negative Square Root: \( x = -\sqrt{10} \)
Isolating Terms
Isolating terms is your first mission when solving an equation, especially quadratics. It's about getting a clear playing field to tackle the variable. Consider the expression: \[ 5x^2 + 1 = 51. \]The goal is to strip it down to the bare essentials. This often means moving everything that isn't attached to nature through addition or subtraction. In this context, you have:- First, subtract 1 from both sides: \[ 5x^2 = 51 - 1 \] This simplifies to: \[ 5x^2 = 50. \]Now, you have successfully isolated the squared term \( 5x^2 \). Keep focusing on getting the variable on one side to make solving a breeze.
  • Look out for constants and move them away.
  • Consider balancing both sides evenly whenever you add or subtract.
Solving Equations
Solving equations is the final thrilling step in the journey to find the unknown variable. Once you have your equation nice and simple, it's time to solve for \( x \). Let's take it further, using this equation as our guide: \[ x^2 = 10 \]Using the square root method, you square root each side: \[ x = \pm \sqrt{10}. \]This gives you two possible solutions for \( x \): \( x = +\sqrt{10} \) AND \( x = -\sqrt{10} \).At the conclusion, you should always reflect on whether your solutions make sense within the context of the problem.
  • Check both solutions in the original equation.
  • Ensure your solutions don鈥檛 result in a contradiction.
Remember, solving equations is about finding balance, and understanding one side must always equal the other.

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Most popular questions from this chapter

Solve each rational inequality in Exercises \(29-48,\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ \frac{x}{x-1}>2 $$

Solve each rational inequality in Exercises \(29-48,\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ \frac{x+3}{x+4}<0 $$

In Exercises \(9-20,\) find each product and write the result in standard form. $$(-5+3 i)(-5-3 i)$$

Which one of the following is true? a. The solution set of \(x^{2}>25\) is \((5, \infty)\) b. The inequality \(\frac{x-2}{x+3}<2\) can be solved by multiplying both sides by \(x+3\), resulting in the equivalent inequality \(x-2<2(x+3)\) c. \((x+3)(x-1) \geq 0\) and \(\frac{x+3}{x-1} \geq 0\) have the same solution set. d. None of these statements is true.

In Exercises \(1-16,\) solve and check each linear equation. $$ 7 x-5=72 $$
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