I don't know how many boys are in the classroom?
Length of room, l = 3x + 1 ft.
Breadth of room, b = x² -1 ft.
Now, it is given that university wants each room to have 195 ft² of living space.
So,
So, above equation has two complex root and one real root i.e x = 4.08 ft .
Therefore, Length of room is 13.24 ft and breadth is 15.65 ft.
Hence, this is the required solution.
Umm could you maybe post a picture of the question?
I have a solution here for the same problem but a different given:
<span>(2x-82)/(x^2+2x-48) dx
</span>
In which this solution will help you answer the problem by yourself:
(2x - 82) /(x² + 2x - 48)
<span>first, factor the denominator completely: </span>
<span>(x² + 2x - 48) = </span>
<span>(x² + 8x - 6x - 48) = </span>
<span>x(x + 8) - 6(x + 8) = </span>
<span>(x - 6)(x + 8) </span>
<span>so the function becomes: </span>
<span>(2x - 82) /[(x - 6)(x + 8)] </span>
<span>let's decompose this into partial fractions: </span>
<span>(2x - 82) /[(x - 6)(x + 8)] = A/(x - 6) + B/(x + 8) </span>
<span>(letting [(x - 6)(x + 8)] be the common denominator at the right side too) </span>
<span>(2x - 82) /[(x - 6)(x + 8)] = [A(x + 8) + B(x - 6)] /[(x - 6)(x + 8)] </span>
<span>(equating numerators) </span>
<span>2x - 82 = Ax + 8A + Bx - 6B </span>
<span>2x - 82 = (A + B)x + (8A - 6B) </span>
<span>yielding the system: </span>
<span>A + B = 2 </span>
<span>8A - 6B = - 82 </span>
<span>A = 2 - B </span>
<span>4A - 3B = - 41 </span>
<span>A = 2 - B </span>
<span>4(2 - B) - 3B = - 41 </span>
<span>A = 2 - B </span>
<span>8 - 4B - 3B = - 41 </span>
<span>A = 2 - B </span>
<span>- 7B = - 41 - 8 </span>
<span>A = 2 - B </span>
<span>7B = 49 </span>
<span>A = 2 - 7 = - 5 </span>
<span>B = 49/7 = 7 </span>
<span>hence: </span>
<span>(2x - 82) /[(x - 6)(x + 8)] = A/(x - 6) + B/(x + 8) = - 5/(x - 6) + 7/(x + 8) </span>
<span>thus the answer is: </span>
<span>(2x - 82) /(x² + 2x - 48) = [- 5 /(x - 6)] + [7 /(x + 8)] </span>
