Question 120·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

Type $t=-6$ and $y=x^4-2tx^2+t^2-9$ in Desmos. An $x$-intercept is where the curve meets the $x$-axis, making $y=0$ and solving the given equation. Desmos shows the curve's lowest point at $(0,27)$, above the axis, so this value gives no real solutions.

Change the first line to $t=3$ and leave the graph equation in place. Desmos shows intercepts at about $x=-2.449$, $x=0$, and $x=2.449$. The curve touches the axis at $0$ without crossing, but $0$ still solves the equation. This value gives three distinct real solutions, not two.

Change the first line to $t=8$. The graph crosses the $x$-axis near $x=-3.317$, $-2.236$, $2.236$, and $3.317$. These are four different intercepts, so this value gives four distinct real solutions.

Change the first line to $t=-2$. Desmos shows intercepts at $(-1,0)$ and $(1,0)$, so the equation has exactly two distinct real solutions, $x=-1$ and $x=1$. The required value of $t$ is $-2$. Choice B.

The first three terms form a perfect square because $x^4=(x^2)^2$. Rewrite the given equation:

$(x^2-t)^2-9=0$. The fourth power is part of a square in $x^2$.

Since $9=3^2$, use the difference of squares pattern, $A^2-B^2=(A-B)(A+B)$: $(x^2-t-3)(x^2-t+3)=0$. This gives two smaller equations to check instead of one fourth-degree equation.

By the zero-product property, a product is $0$ only if at least one factor is $0$: $x^2-t-3=0\quad\text{or}\quad x^2-t+3=0$. Move the constants to the other sides: $x^2=t+3\quad\text{or}\quad x^2=t-3$. Keep both branches when counting solutions.

For $x^2=c$, positive $c$ gives two real roots, zero gives one, and negative $c$ gives none. A real number's square can't be negative, and positive squares come from two opposite numbers. Since $t+3$ is greater than $t-3$, exactly two roots require $t+3>0$ and $t-3<0$: the first branch gives a pair, and the second gives none. If the second equals $0$, it adds $x=0$ as a third root.

Solve the condition that gives the pair of roots. Subtract $3$ from $t+3>0$: $t>-3$. So $t$ must be greater than $-3$.

Solve the condition that gives no extra roots. Add $3$ to $t-3<0$: $t<3$. Together with the lower bound, this means $-3<t<3$.

Of the offered values, only $-2$ lies strictly between $-3$ and $3$. It makes one equation for $x^2$ positive and the other negative, giving exactly two distinct real solutions. Choice B.