length of the rectangle is five more than twice its width .area of the rectangle is 52 sq meter. find the length of the rectangle
have my own sol. pls. continue
L=5+2w
LW=52
(5+2w)w=52
2w^2+5w=52
2w^2+5w-52=0
soo. pls could someone continue this~?
your equation factors to
(w-4)(2w + 13) = 0
so w = 4 or w = -13/2, which makes no sense
so the width is 4 and the length is 2(4) + 5 or 13
2w^2+5w-52=0.00
2w^2+5w-52=0
To continue solving the equation, you can use the quadratic formula:
Given the quadratic equation: 2w^2 + 5w - 52 = 0
Using the quadratic formula:
w = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 2, b = 5, and c = -52.
Substituting these values into the quadratic formula, we get:
w = (-5 ± √(5^2 - 4*2*(-52))) / 2*2
Simplifying further:
w = (-5 ± √(25 + 416)) / 4
w = (-5 ± √441) / 4
Now, let's solve for w:
Case 1: w = (-5 + √441) / 4
w = (-5 + 21) / 4
w = 16 / 4
w = 4
Case 2: w = (-5 - √441) / 4
w = (-5 - 21) / 4
w = -26 / 4
w = -6.5
Since the width of a rectangle cannot be negative, we discard the negative value of w.
Therefore, the width of the rectangle (w) is 4.
Now, to find the length (L) of the rectangle, we can substitute the value of w into the equation: L = 5 + 2w
L = 5 + 2(4)
L = 5 + 8
L = 13
So, the length of the rectangle is 13 meters.