Solve the equation. $10 ( m + 3 ) ^ { - 2 } - 29 ( m + 3 ) ^ { - 1 } + 9 = 0$
A) $\frac { 29 \pm \sqrt { 481 } } { 18 } - 3$
B) $\frac { 29 \pm \sqrt { 481 } } { 18 } + 3$
C) $$\frac { 29 \pm \sqrt { 481 } } { 20 } + 3$$
D) $\frac { 29 \pm \sqrt { 481 } } { 20 } - 3$

Question

Quizplus · Accepted Answer

This equation can be solved by substituting $y = m + 3$ to transform it into a quadratic equation in terms of $y$: $10y^{-2} - 29y^{-1} + 9 = 0$. Multiplying through by $y^2$ gives a standard quadratic form: $10 - 29y + 9y^2 = 0$, or $9y^2 - 29y + 10 = 0$. Solving this quadratic equation for $y$ using the quadratic formula, $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 9$, $b = -29$, and $c = 10$, gives $y = \frac{29 \pm \sqrt{(-29)^2 - 4(9)(10)}}{2(9)} = \frac{29 \pm \sqrt{841 - 360}}{18} = \frac{29 \pm \sqrt{481}}{18}$. Substituting back $y = m + 3$ gives $m + 3 = \frac{29 \pm \sqrt{481}}{18}$, so $m = \frac{29 \pm \sqrt{481}}{18} - 3$, which matches choice A.

Solve the Equation 10(m+3)−2−29(m+3)−1+9=010 ( m + 3 ) ^ { - 2 } - 29 ( m + 3 ) ^ { - 1 } + 9 = 010(m+3)−2−29(m+3)−1+9=0

Solve the Equation $10 ( m + 3 ) ^ { - 2 } - 29 ( m + 3 ) ^ { - 1 } + 9 = 0$