Super-Hard SAT Math Question 153 | Free SAT Prep | Aniko

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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 $x$ ? аnікo.аi/‌sаt

For a line and a parabola, intersections come from setting the two expressions for $y$ equal, which creates a quadratic in $x$ . “Exactly one intersection” means that quadratic has one real solution, so set the discriminant to 0 to determine the parameter (here, $p$ ). Use any given restriction on the parameter (such as $p<0$ ) to choose the valid case, then substitute back to find the single (double) root for $x$ . аniкo.‍аi/ѕаt

Hints

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Turn it into one equation in one variable

At the intersection point, both expressions for $y$ are equal. Set the right-hand sides equal and simplify.

Use the meaning of “exactly one point”

After simplifying, you should have a quadratic in $x$ . A quadratic has exactly one real solution when its discriminant equals 0.

Don’t ignore the sign of $p$

Solving the discriminant equation will likely give two possible values of $p$ . Use the fact that $p$ is negative to choose the right one. © аnіkο‌.ai

Desmos Guide

Graph the parabola and a parameterized line

Enter the parabola: y=2x^2-7x+60.

Then enter the line using a slider: y=p x+10 (Desmos will create a slider for $p$ ). Frоm аn⁠і⁠кο.аi

Adjust $p$ so there is exactly one intersection and $p$ is negative

Move the slider so the line is tangent to the parabola (you should see the graphs touch at exactly one point).

Make sure the slider value stays below 0.

Read the $x$ -coordinate and match it to a choice

Click the single intersection point and look at its $x$ -coordinate. Select the answer choice that matches that $x$ -value.

Step-by-step Explanation

Set the equations equal to get a quadratic in $x$

At an intersection point, the $y$ -values are equal:

2x^2 - 7x + 60 = px + 10

Rearrange to standard form:

2x^2 + (-7-p)x + 50 = 0

Use the “exactly one intersection” condition

A quadratic has exactly one real solution when its discriminant is 0.

Here, $a=2$ , $b=-7-p$ , and $c=50$ , so

\begin{aligned} 0 &= b^2 - 4ac \\ &= (-7-p)^2 - 4(2)(50) \\ &= (p+7)^2 - 400 \end{aligned}

So $(p+7)^2=400$ , which gives $p+7=20$ or $p+7=-20$ , so $p=13$ or $p=-27$ .

Because $p$ is negative, it must be $p=-27$ . Powerеd bу А⁠nіко

Substitute $p$ to find the (double) solution for $x$

Substitute $p=-27$ into the quadratic:

2x^2 + (-7-(-27))x + 50 = 0

2x^2 + 20x + 50 = 0

Divide by 2:

x^2 + 10x + 25 = 0

Factor:

(x+5)^2 = 0

State the value of $x$

From $(x+5)^2=0$ , the (single) solution is $x=-5$ .

Therefore, the correct choice is $-5$ .

Strategy

Hints

Desmos Guide

Step-by-step Explanation

Study Hints

Strategy

Hints

Desmos Guide

Step-by-step Explanation