Question 153·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

Type $t=0$, then both given equations in Desmos. Zero is the smallest allowed value of $t$, and the graphs cross twice there. Each crossing solves both equations, so the goal is to find when those crossings disappear.

At a shared point, $y=3x+t$. Use substitution to replace every $y$ in the first equation, including the squared one: $x^2+(3x+t)^2-6x-8(3x+t)+9=0$.

Expand $(3x+t)^2$ and distribute $-8$. Keep both terms of the squared expression: $x^2+9x^2+6tx+t^2-6x-24x-8t+9=0$.

Combine like terms so you can identify the quadratic’s coefficients: $10x^2+(6t-30)x+(t^2-8t+9)=0$. Each real value of $x$ here gives a shared point, since the line then supplies its $y$-value.

The discriminant of $ax^2+bx+c=0$ is $D=b^2-4ac$. Here $a=10$, $b=6t-30$, and $c=t^2-8t+9$, so: $D=(6t-30)^2-4(10)(t^2-8t+9)$.

Expand both products: $D=36t^2-360t+900-40t^2+320t-360$. Combine like terms: $D=-4t^2-40t+540$. The sign of $D$, rather than a solution for $x$, is what you need.

Factor out $4$: $D=4(135-t^2-10t)$. Since $(t+5)^2=t^2+10t+25$, replace $t^2+10t$ to make the boundary easier to find: $D=4[160-(t+5)^2]$. This completed square shows how $D$ changes as $t$ increases.

Zero intersections requires a negative discriminant; a zero discriminant still gives one touching point. Set $D<0$: $4[160-(t+5)^2]<0$. Divide by the positive $4$: $160-(t+5)^2<0$. Move the square to the other side: $(t+5)^2>160$.

Because $t$ is nonnegative, $t+5$ is positive. Take the positive square root: $t+5>\sqrt{160}$. Since $\sqrt{160}=4\sqrt{10}$, subtract $5$: $t>4\sqrt{10}-5$.

Type $4\sqrt{10}-5$ on a new Desmos line. It shows $7.649\ldots$. The least integer strictly above that cutoff is $8$, so that is the least possible value of $t$ with no real solution. Grid in 8.