Quadratic equation

[airplanes123]

I have to make a quadratic equation and then solve it
problem is the perimeter of a rectangle is 34 cm and diagonal is 13 cm. determine lengths of the sides of the rectangle.
2x+2y=34
2y=-2x+34
y=-1x+17
or is it
x^2 + y^2=13^2
sincerely,
confused. :/

 

[mmm4444bot]

It is always good form to state definitions for all symbols that you choose, when starting a word problem:

Let x = one dimension of the rectangle

Let y = the other dimension of the rectangle

2x + 2y = 34
2y = -2x + 34
y = -1x + 17

We do not write coefficients of 1 (highlighted in red above).

Write y = -x + 17 instead.


or is it
x^2 + y^2 = 13^2

You can use both equations together.

You have arrived at the following "system of two equations" in x and y:

y = -x + 17

x^2 + y^2 = 169

To solve for x, you could now substitute the expression -x + 17 for y in the second equation:

x^2 + (-x + 17)^2 = 169 ? This is a quadratic equation

Expand the square and move all terms to one side, and you'll have the familiar form Ax^2 + Bx + C = 0.

Solve that for x (you will get two values).

Substitute each of these into your earlier result for y:

y = -x + 17

You should get the same two values for y as you did for x. This suggests that either of the two solutions for x could be the length of the rectangle with the other value the width.

:idea: Check your results, by confirming that your dimensions for length and width jive with the given perimeter and diagonal information.