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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 11: Q. 71 (page 774)

The difference of two numbers is $2$ , and the sum of their
squares is $10$ . Find the numbers.

Short Answer

Expert verified

The numbers are $3$ and $1$ or $- 1$ and $- 3$ .

Step by step solution

01

Step 1. Given information

Given that,

The difference of the two number is $2$ . So, $a - b = 2$ .

The sum of their squares is $10$ . So, $a^{2} + b^{2} = 10$ .

So, the system of equations are,

$\{\begin{cases} a - b = 2 \\ a^{2} + b^{2} = 10 \end{cases}$

02

Step 2 Let us write the first equation as,

$a - b = 2 a = b + 2$

Substitute the value of $a$ in the first equation.

${(b + 2)}^{2} + b^{2} = 10 b^{2} + 4 b + 4 + b^{2} = 10 2 b^{2} + 4 b + 4 - 10 = 0 2 b^{2} + 4 b - 6 = 0$

Now factor out the common terms.

$2 (b^{2} + 2 b - 3) = 0 b^{2} + 2 b - 3 = 0$

03

Step 3 we can solve for b by using the quadratic formula -b±b2-4ac2a where a=1,b=2,c=-3.

$b = \frac{- 2 卤 \sqrt{2^{2} - 4 (1) (- 3)}}{2 (1)} b = \frac{- 2 卤 4}{2} b = \frac{- 2 + 4}{2} o r b = \frac{- 2 - 4}{2} b = \frac{2}{2} o r b = \frac{- 6}{2} b = 1 o r b = - 3$

04

Step 4 Substitute the value of b in the first equation.

Put $b = 1$ in the first equation.

$a - 1 = 2 a = 2 + 1 a = 3$

Put $b = - 3$ in the first equation.

$a - (- 3) = 2 a + 3 = 2 a = 2 - 3 a = - 1$

So, the required numbers are, $3$ and $1$ or $- 1$ and $- 3$ .

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Most popular questions from this chapter

In Problems 23鈥�34, graph each system of linear inequalities by hand. Verify your results using a graphing utility

$\{\begin{cases} 2 x + y 鈮� - 2 \\ 2 x + y 鈮� 2 \end{cases}$

The difference of two numbers is the same as their product, and the sum of their reciprocals is $5$ . Find the numbers.

Verify that the values of the variables listed are solutions of the system of equations.

$\{\begin{matrix} \begin{array}{rcl} 3 x + 3 y + 2 z & = & 4 \\ x - 3 y + z & = & 10 \end{array} \\ 5 x - 2 y - 3 z = 8 \end{matrix} x = 2, y = - 2, z = 2; (2, - 2, 2)$

Verify that the values of the variables listed are solutions of the system of equations.

$\{\begin{matrix} x - y = 3 \\ \frac{1}{2} x + y = 3 \end{matrix} x = 4, y = 1; (4, 1)$

Solve each system of equations. If the system has no solution, say that it is inconsistent.

$\{\begin{cases} 2 x + 4 y = \frac{2}{3} \\ 3 x - 5 y = - 1 \end{cases}$

See all solutions

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