Needing help with this math question

Question

The width of a rectangle is 9 less than twice its length. If the area of the rectangle is 135 , what is the length of the diagonal?

Answer 1

This should be solved through quadratic equation formula ===>

b +/- SqRT( b^2 - 4ac) / 2a.

Let x be length of rectangle.

Then width is (2x - 9)

Area of rectangle then becomes x(2x - 9) = 135

==> 2x^2 -9x - 135 = 0

With Formula for quadratic equation

x = ( -9 +/- SqRT( 81 - 4x2x-135) )/ 2x2

width = (2x -9) ===> (-9 +/- SqRT( 81 - 4x2x-135)) /2 - 9

diagonal = SqRT( x^2 + (2x -9)^2)

===> SqRT ( ( ( -9 +/- SqRT( 81 - 4x2x-135))/4)^2 + (( -9 +/- SqRT(81 - 4x2x-135))/2 )^2 )

Let (-9 +/- SqRT(81 - 4x2x-135) ) / 2 be a.

then diagonal is ==>

SqRT ( (a/2)^2 + a^2) ) = SqRT( a^2/4 +a^2) = 1/2 x SqRT(2)a

Solve.

Answer 2

Given: w = 2L - 9, A = 135 cm²

The formula for the area of a rectangle is A = L*w, so plug in all the given in this formula, you'll have:

135 = L *(2L - 9)

135 = 2L² - 9L

0 = 2L² - 9L -135

This is a quadratic equation and one way to solve that is by using quadratic formula and will get two answers:

L₁ = 10.7683625187 and L₂ = -6.2683625187

We eliminate L₂ because we can't have a negative length so L = 10.7683625187.

For the width, following the given formula above, we have:

w = 2(10.7683625187) - 9

w = 12.5367250374

Since the diagonal, length and width form a right triangle, we can use the Pythagorean Theorem to solve for the diagonal:

d² = (12.5367250374)² + (10.7683625187)²

d ≈ 16.53 cm