Question 153·200 Super-Hard SAT Math Questions·Advanced Math

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y = 2x^2 - 7x + 60

y = px + 10

In the given system of equations, $p$ is a negative constant. The graphs of the equations in the system intersect at exactly one point, $(x, y)$ , in the $xy$ -plane. Which choice could be the value of $x$ ?

A line that meets a parabola once is tangent: it touches at one point. Set their expressions for $y$ equal, then make the resulting quadratic’s discriminant, the number under the square root in the quadratic formula, equal zero. Apply the constant’s sign condition before graphing the resulting line and parabola in Desmos to read the requested coordinate.

Hints

Hint 1
At an intersection, both graphs have the same $y$ for the same $x$ . Set the two expressions for $y$ equal, then move everything to one side to get a quadratic in $x$ .
Hint 2
A quadratic has one real solution when its discriminant, $b^2-4ac$ , is zero: the plus and minus parts of the quadratic formula give the same value. What are $a$ , $b$ , and $c$ after you set the expressions equal?
Hint 3
A squared equation can allow two values for a constant. Use the condition that $p$ is negative when finding it; then graph the resulting line to read the intersection’s $x$ -coordinate.

Step-by-step

Find the tangent line, then read the intersection
Step 1
Make one equation for the shared $x$
At an intersection, the two $y$ -values agree, so set the right sides equal:
$2x^2-7x+60=px+10.$
Step 2
Put the equation in quadratic form
Subtract $px+10$ from both sides to get a quadratic, an equation with $x^2$ :
$2x^2-(p+7)x+50=0.$
Step 3
Use the one-intersection condition
For $ax^2+bx+c=0$ , the discriminant is $b^2-4ac$ . Exactly one intersection means the resulting quadratic has a zero discriminant. When it’s zero, the plus and minus parts of the quadratic formula give the same $x$ . Here $a=2$ , $b=-(p+7)$ , and $c=50$ , so set the discriminant to zero:
$[-(p+7)]^2-4(2)(50)=0.$
Step 4
Calculate the discriminant’s constant
Type $4(2)(50)$ in Desmos; it prints $400$ . Add $400$ to both sides of the discriminant equation:
$(p+7)^2=400.$
Step 5
Find the negative value of $p$
Taking square roots gives $p+7=\pm\sqrt{400}$ . The positive branch gives $p=-7+20=13$ , which breaks the negative condition. For the negative branch, subtract $7$ :
$p=-7-\sqrt{400}.$
Type $-7-\sqrt{400}$ in Desmos; it prints $-27$ , the negative value of $p$ .
Step 6
Read the requested coordinate
Type $y=2x^2-7x+60$ and $y=-27x+10$ , replacing $p$ with the value found. Click their one shared point; Desmos shows $(-5,145)$ . Its first coordinate is $x$ , so the requested value is $-5$ . Choice A.

Question 153·200 Super-Hard SAT Math Questions·Advanced Math

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