Question

the square of one number is 2 less than twice the square of another. the sum of the squares of the two numbers is 25. find the numbers. help please.

Answer 1

Choose two variables to represent the two numbers.

Answer 2

can i use x and y?

Answer 3

Of course.

Answer 4

Now let the first number be x. How do you express "the square of one number" using x?

Answer 5

the square of one number is 2 less than twice the square of another
\(x^2\) = \(2y^2 - 2\)

Answer 6

From the first statement, we have the equation:
\(x^2 = 2y^2 - 2\)

The second sentence is:
"the sum of the squares of the two numbers is 25."

Now we need to translate the second sentence into an equation.
Can you try doing it?

Answer 7

\[x ^{2}+2y ^{2}=25?\]

Answer 8

No, close, but there is no 2.

Just
\(x^2 + y^2 = 25\)

Answer 9

Now we have two equations that we need to solve as a system of equations:

\(x^2 = 2y^2 - 2\)
\(x^2 + y^2 = 25\)

Let's solve the second equation for \(x^2\).

\(x^2 = 25 - y^2\)

Ok so far?

Answer 10

yes

Answer 11

Since we know \(x^2 = 2y^2 - 2\), and \(x^2 = 25 - y^2\)
then that means
\(2y^2 - 2 = 25 - y^2\)

Add 2 to both sides, and add \(y^2\) to both sides.

\(3y^2 = 27\)

Divide both sides by 3

\(y^2 = 9\)

Answer 12

ok i understand that

Answer 13

Now we can find y.

Answer 14

In general, if you have an equation of the form
\(x^2 = k\),
where k is a number, then the solution is
\(x = \sqrt{k}\) or \(x = -\sqrt{k} \)

Answer 15

In this case, we have
\(y^2 = 9\), so we get
\(y = \sqrt{9}\) or \(y = -\sqrt{9} \)
which gives us
\( y = 3\) or \(y = -3\)

Answer 16

Now that we have a solution for y, we plug the values of y into one of our equations to get x. We need to do it for both values of y.
Let's us the second equation:
\(x^2 + y^2 = 25\)

Answer 17

Let's start with y = 3:
\(x^2 + 3^2 = 25\)
\(x^2 + 9 = 25\)
\(x^2 = 16\)
\(x = 4\) or \(x = -4\)
This means that for y = 3, we have two solutions to the system of equations:
x = 4, y = 3 and x = -4, y = 3

Answer 18

Now we do the same for y = -3:
\(x^2 + (-3)^2 = 25\)
\(x^2 + 9 = 25\)
\(x^2 = 16\)
\(x = 4\) or \(x = -4\)
This means that for y = -3, we have two solutions to the system of equations:
x = 4, y = -3 and x = -4, y = -3

Answer 19

There are 4 pairs of numbers that work, so there are 4 answers to the question:
3, 4
3, -4
-3, 4
-3, -4

Answer 20

will there always be a positive and negative solution?

Answer 21

Not always, but when you deal with squares, sometimes you do get them.

Answer 22

thank you!

Answer 23

wlcm