Question

-9x(x + 9) = 90
one solution to the given equation can be written as ( x = \frac{-s pm sqrt{t}}{2} ), where ( s ) and ( t ) are positive integers. what is the value of ( \frac{s}{t} )?

Explanation:

Step1: Expand left-hand side

$-9x(x+9) = -9x^2 -81x$
Equation becomes: $-9x^2 -81x = 90$

Step2: Rearrange to standard quadratic form

Bring all terms to one side: $-9x^2 -81x -90 = 0$
Divide by $-9$: $x^2 +9x +10 = 0$

Step3: Apply quadratic formula

For $ax^2+bx+c=0$, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Here $a=1, b=9, c=10$:
$x=\frac{-9\pm\sqrt{9^2-4(1)(10)}}{2(1)}=\frac{-9\pm\sqrt{81-40}}{2}=\frac{-9\pm\sqrt{41}}{2}$

Step4: Identify s and t, compute ratio

$s=9$, $t=5$ (from $\sqrt{41}$, the form is $\frac{-s\pm\sqrt{...}}{t}$)
$\frac{s}{t}=\frac{9}{5}$

Answer:

$\frac{9}{5}$