Decide what values of the variable cannot possibly be solutions for the equation.
-$$25 x ^ { 4 } - 61 x ^ { 2 } + 36 = 0$$
A) $\left\{ - 1 , - \frac { 5 } { 6 } \right\}$

B) $\left\{ - \frac { 6 } { 5 } , - 1,1 , \frac { 6 } { 5 } \right\}$

C) $\left\{ - 1 , - \frac { 5 } { 6 } , \frac { 5 } { 6 } , 1 \right\}$

D) $\left\{ 1 , \frac { 6 } { 5 } \right\}$

Question

Decide what values of the variable cannot possibly be solutions for the equation.
-$$25 x ^ { 4 } - 61 x ^ { 2 } + 36 = 0$$
A) $\left\{ - 1 , - \frac { 5 } { 6 } \right\}$

B) $\left\{ - \frac { 6 } { 5 } , - 1,1 , \frac { 6 } { 5 } \right\}$

C) $\left\{ - 1 , - \frac { 5 } { 6 } , \frac { 5 } { 6 } , 1 \right\}$

D) $\left\{ 1 , \frac { 6 } { 5 } \right\}$

Quizplus · Accepted Answer

This equation is a quartic equation in terms of $x^2$, not $x$. Let $y = x^2$. The equation becomes $25y^2 - 61y + 36 = 0$, which can be solved for $y$. The solutions for $y$ will be positive because $y = x^2$ and a square is always non-negative. However, when we take the square root of $y$ to solve for $x$, each positive $y$ value gives two $x$ values: one positive and one negative. Therefore, the original equation can have solutions for both positive and negative values of $x$, making it incorrect to exclude any real number based solely on its sign. The correct answer involves recognizing that the equation, when solved, will yield specific values for $x^2$, and thus specific values for $x$, both positive and negative.

Decide What Values of the Variable Cannot Possibly Be Solutions 25x4−61x2+36=025 x ^ { 4 } - 61 x ^ { 2 } + 36 = 025x4−61x2+36=0

Decide What Values of the Variable Cannot Possibly Be Solutions $25 x ^ { 4 } - 61 x ^ { 2 } + 36 = 0$