Answer :
Since the equation for area of a rectangle is A = l×w, and the length is 5 more than twice the width, making it 2w + 5, you have to plug in the quantities.
A = l×w
12 = (2w+5)×w
12 = 2w²+5w
0 = 2w²+5w-12
0 = (2w-3)(w+4)
Since the width can't be a negative number, the answer is:
2w-3 = 0
2w = 3
w = 3/2
And now, since length is 2w+5, you plug in the value of the width into the equation.
2(3/2)+5
3+5
8
A = l×w
12 = (2w+5)×w
12 = 2w²+5w
0 = 2w²+5w-12
0 = (2w-3)(w+4)
Since the width can't be a negative number, the answer is:
2w-3 = 0
2w = 3
w = 3/2
And now, since length is 2w+5, you plug in the value of the width into the equation.
2(3/2)+5
3+5
8
Let the width of the triangle be x.
Therefore length = 2x + 5. (5 more than twice its width).
Area = L * W = (2x + 5) x = 12.
2x^2 + 5x = 12
2x^2 + 5x - 12 = 0.
This is a quadratic equation. (2)*(-12) = -24.
We think of two numbers whose product is -24 and it sum is +5.
Those two numbers are 8 and -3. So we replace the middle term of
quadratic with (8 -3).
2x^2 + 5x - 12 = 0.
2x^2 + 8x-3x - 12 = 0. Factorize.
2x(x + 4) - 3(x +4) = 0
(2x-3)(x+4) = 0.
(2x-3) = 0 or (x+4) = 0
2x = 3. x = 0 -4
x = 3/2 = 1.5 x = -4. (x can't be negative, since we are solving for lengths)
x = 1.5 is only valid solution.
width = x = 1.5
Length = (2x + 5) = 2*1.5 +5 = 3 + 5 =8.
Therefore length = 8, and width = 1.5
Cheers.
Therefore length = 2x + 5. (5 more than twice its width).
Area = L * W = (2x + 5) x = 12.
2x^2 + 5x = 12
2x^2 + 5x - 12 = 0.
This is a quadratic equation. (2)*(-12) = -24.
We think of two numbers whose product is -24 and it sum is +5.
Those two numbers are 8 and -3. So we replace the middle term of
quadratic with (8 -3).
2x^2 + 5x - 12 = 0.
2x^2 + 8x-3x - 12 = 0. Factorize.
2x(x + 4) - 3(x +4) = 0
(2x-3)(x+4) = 0.
(2x-3) = 0 or (x+4) = 0
2x = 3. x = 0 -4
x = 3/2 = 1.5 x = -4. (x can't be negative, since we are solving for lengths)
x = 1.5 is only valid solution.
width = x = 1.5
Length = (2x + 5) = 2*1.5 +5 = 3 + 5 =8.
Therefore length = 8, and width = 1.5
Cheers.
