Question

The area of a rectangle is given by the equation y=x^2-9x+14. If the area is 50 square units, what is the value of x?

Answer 1

`\bf \textit{that simply means }A=y\implies 50=x^2-9x+14 \\\\\\ 0=x^2-9x+14-50\implies \begin{array}{lccll} x^2&-9x&-36\\ &\uparrow &\uparrow \\ &-12+3&-12\cdot 3 \end{array} \\\\\\ 0=(x-12)(x+3)\implies \begin{cases} 0=x-12\implies &\boxed{12=x}\\ 0=x+3\implies &-3=x \end{cases}`

since the value for "x" cannot be negative, because is a length unit, so is not -3.

Answer 2

50 = x^2 - 9x + 14
x^2 - 9x + 14 - 50 = 0
x^2 - 9x - 36 = 0
(x - 12)(x + 3) = 0

x - 12 = 0
x = 12 <==

x + 3 = 0
x = -3......not this one