Super-Hard SAT Math Question 23 | Free SAT Prep | Aniko

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Question 23·200 Super-Hard SAT Math Questions·Advanced Math

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For the system of equations below, $k$ is a constant. If the system has exactly one real solution $(x,y)$ and $k>0$ , which choice gives the value of $k$ ?

\begin{cases} x^2+y^2=12x \\ y=x+k \end{cases}

When a system mixes a linear equation with a nonlinear equation (like a circle), substitute the linear expression into the nonlinear equation to get a single-variable quadratic. The phrase “exactly one real solution” is a big clue to use the discriminant: set $b^2-4ac=0$ so the quadratic has a repeated root, then apply any given condition (here, $k>0$ ) to choose the correct value.

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Turn the system into one equation in one variable

Use $y=x+k$ to replace $y$ in the equation $x^2+y^2=12x$ .

Watch what “exactly one real solution” implies

After substitution you should get a quadratic in $x$ . A quadratic has exactly one real solution when it has a repeated root.

Use the discriminant

For $ax^2+bx+c=0$ , a repeated root happens when $b^2-4ac=0$ . Set that equal to 0 and solve for $k$ , then use $k>0$ .

Desmos Guide

Graph the circle in standard form

Rewrite $x^2+y^2=12x$ as $(x-6)^2+y^2=36$ , then enter:

(x-6)^2 + y^2 = 36

Graph the line with a slider for $k$

Enter:

y = x + k

Desmos will create a slider for $k$ .

Adjust $k$ until there is exactly one intersection

Move the slider until the line just touches the circle (tangent), meaning there is exactly one intersection point shown.

Match the slider value to a choice

At tangency, the slider value is about $2.49$ . The only option equal to approximately $2.49$ is $-6+6\sqrt{2}$ .

Step-by-step Explanation

Substitute to get a quadratic in one variable

Substitute $y=x+k$ into $x^2+y^2=12x$ :

\begin{aligned} x^2+(x+k)^2&=12x \\ 2x^2+2kx+k^2&=12x \\ 2x^2+(2k-12)x+k^2&=0 \end{aligned}

This is a quadratic equation in $x$ .

Use the “exactly one real solution” condition

A quadratic $ax^2+bx+c=0$ has exactly one real solution when its discriminant is 0:

b^2-4ac=0

Here, $a=2$ , $b=2k-12$ , and $c=k^2$ , so:

(2k-12)^2-4(2)(k^2)=0

Solve for $k$ and apply $k>0$

Simplify and solve:

\begin{aligned} (2k-12)^2-8k^2&=0 \\ 4(k-6)^2-8k^2&=0 \\ (k-6)^2-2k^2&=0 \\ (k^2-12k+36)-2k^2&=0 \\ -k^2-12k+36&=0 \\ k^2+12k-36&=0 \end{aligned}

\begin{aligned} k&=\frac{-12\pm\sqrt{12^2-4(1)(-36)}}{2} \\ &=\frac{-12\pm\sqrt{144+144}}{2} \\ &=\frac{-12\pm 12\sqrt{2}}{2} \\ &=-6\pm 6\sqrt{2} \end{aligned}

Since $k>0$ , the value is $-6+6\sqrt{2}$ . That corresponds to $-6+6\sqrt{2}$ .

Strategy

Hints

Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation