A rectangle with an area of 24 square units has a length of [tex]x + 1[/tex] and a width of [tex]4x - 6[/tex]. Find the value of [tex]x[/tex].
Answer (3)
a re a = l e n g t h ⋅ w i d t h A = 24 s 2 l = x + 1 , w = 4 x − 6 A = lw 24 = ( x + 1 ) ( 4 x − 6 )
24 = 4 x 2 − 6 x + 4 x − 6 4 x 2 − 6 x + 4 x − 6 − 24 = 0 4 x 2 − 2 x − 30 = 0 / : 2 2 x 2 − x − 15 = 0 a = 2 , b = − 1 , c = − 15
Δ = b 2 − 4 a c = ( − 1 ) 2 − 4 ⋅ 2 ⋅ ( − 15 ) = 1 + 120 = 121 x 1 = 2 a − b − Δ = 2 ∗ 2 1 − 121 = 4 1 − 11 = 4 − 10 = − 2 5 x 2 = 2 a − b + Δ = 2 ∗ 2 1 + 121 = 4 1 + 11 = 4 12 = 3 U se p os i t i v e v a l u e o f x : x = 3
To find the value of x for the rectangle with given dimensions and area, we set up and factor a quadratic equation. The valid solution for the rectangle's width to be positive is x=3. We confirm this by substituting back into the original area equation to get 24=24.
To find the value of x for a rectangle with an area of 24 square units, where the length is x+1 and the width is 4x-6, we set up the equation for the area of a rectangle (length × width), which is (x+1)(4x-6) = 24. We then expand this to form a quadratic equation: 4x^2 -6x + 4x -6 = 24, which simplifies to 4x^2 - 2x - 30 = 0.
Next, we factor the quadratic equation to find the values of x that satisfy it. Dividing through by 2, we get 2x^2 - x - 15 = 0. Factoring further, we obtain (2x+5)(x-3) = 0. This gives two possible solutions for x, which are x = 3 and x = -2.5. However, since width cannot be negative in a real-world context, the only valid solution is x=3.
As a quick check, we substitute x = 3 back into the original equation for the area and find that it leads to an identity: 24 = 24, confirming the solution is correct.
The value of x in the area equation for the rectangle is 3. This is obtained through the quadratic formula after setting the area equation in standard form. The process involves expanding the area equation, rearranging it, and solving the quadratic equation for x .
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