Question

I need help? Lol fr tho I forgot.

Answer 1

Answer:

Width of the rectangle is approximately 3.78 yd.

Step-by-step explanation:

Given:

Area of rectangle = $52\ yd^2$

Let the width of the rectangle be 'x'.

Now Given:

Length of the rectangle is 10 more than width.

Length of the rectangle = $x+10$

We need to find the width of the rectangle.

Solution:

Now we know that;

Area of rectangle is given by length times width.

framing in equation form we get;

$x(x+10) =52\\\\x^2+10x-52=0$

Now we will solve by using quadratic formula.

$x=\frac{-b\±\sqrt{b^2-4ac}}{2a}$

In Our case a=1, b=10, c =-52

First we will find;

$\sqrt{b^2-4ac} = \sqrt{10^2-4\times 1\times -52}= \sqrt{100+208}= \sqrt{308} = 17.55$

$x=\frac{-b\±\sqrt{b^2-4ac}}{2a}$

$x=\frac{-10\±17.55}{2\times1}= \frac{-10\±17.55}{2}$

Since roots have 2 values one positive and 1 negative we will find it separately.

$x = \frac{-10+17.55}{2} = \frac{7.55}{2} \approx 3.78$

Also;

$x = \frac{-10-17.55}{2} = \frac{-27.55}{2} \approx -13.78$

Now we got 2 roots of x out of which one is positive and one is negative.

Since the width of the rectangle cannot be negative, Hence we will consider the positive value of 'x'.

Hence width of the rectangle is approximately 3.78 yd.