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Chapter 8: Problem 9

\(y^{2}=x-1 ; x=3\)

Short Answer

Expert verified
The solutions are \[ y = \sqrt{2} \] and \[ y = -\sqrt{2} \]

Step by step solution

01

Substitute the given value of x

The equation is given as \[ y^{2}=x-1 \] We are also given that \[ x=3 \] Substitute 3 for x in the equation to get: \[ y^{2}=3-1 \] This simplifies to: \[ y^{2} = 2 \]
02

Solve for y

To solve for y, take the square root of both sides of the equation \[ y^{2} = 2 \] Taking the square root of both sides gives: \[ y = \pm \sqrt{2} \] So, \[ y = \sqrt{2} \] or \[ y = -\sqrt{2} \]

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Key Concepts

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

Substitution in Equations
Substitution in equations is a fundamental algebraic technique. It involves replacing a variable with a given value. This method simplifies the equation and makes it easier to solve.
In our example, we start with the equation \( y^{2}=x-1 \) and the given value \( x=3 \).
By substituting 3 for x, the equation becomes: \[ y^{2}=3-1 \]
which simplifies to \[ y^{2} = 2 \]. Now, we can move on to solving for the variable y.
Solving for Variables
Once we have substituted the known value, the next step is to solve for the variable. In our case, we need to solve for y in the equation \( y^{2} = 2 \).
This means we need to find the value of y that makes the equation true.
To do this, we isolate y by using mathematical operations that leave it alone on one side of the equation. Here, we will take the square root of both sides.
Thus, if \( A = B \), then \[ \text{if} \ y^{2} = 2, \ \text{then} \ y = \pm \sqrt{2} \].
Note that whenever we take the square root of a squared term, we must consider both the positive and negative roots.
Therefore, the solutions for y are \[ y = \sqrt{2} \ \text{or} \ y = -\sqrt{2} \].
Square Roots
Square roots are essential in algebra for solving equations involving squared terms. Taking the square root of a number is the opposite operation of squaring that number.
For example, the square root of 4 is 2 because \( 2^{2} = 4 \). Similarly, the square root of 9 is 3 because \( 3^{2} = 9 \).
In our exercise, we found that \( y^{2} = 2 \). To find y, we took the square root of both sides: \[ y = \pm \sqrt{2} \].
This introduces the concept of both positive and negative roots.
  • The positive root, denoted as \( \sqrt{2} \), is approximately 1.414.
  • The negative root, denoted as \( -\sqrt{2} \), is approximately -1.414.

This simply means there can be two values of y that satisfy the original equation when we take the square root.

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Most popular questions from this chapter

A square plate of side \(4 \mathrm{ft}\) is submerged vertically in a tank of water and its center is \(2 \mathrm{ft}\) below the surface. Find the force due to liquid pressure on one side of the plate.

\(y=\sqrt{x} ; y=x^{3}\)

The length of a rod is \(L \mathrm{ft}\) and the center of mass of the rod is at the point \(\frac{3}{4} L \mathrm{ft}\) from the left end. If the measure of the linear density at a point is proportional to a power of the measure of the distance of the point from the left end and the linear density at the right end is 20 slugs/ft, find the linear density at a point \(x \mathrm{ft}\) from the left end. Assume the mass is measured in slugs.

Find the area of the region bounded by the three curves \(y=x^{2}, x=y^{3}\), and \(x+y=2\).

\(x=y^{2}-y ; x=y-y^{2}\)
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