SAT Nonlinear Functions Question 191 | Free SAT Prep | Aniko

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Question 191·Medium·Nonlinear Functions

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What is the sum of all distinct real solutions to the equation

x^2 - 5x = 6?

Your answer

(Express the answer as an integer)

For quadratic equations set equal to a constant, first move all terms to one side so the equation equals 0, then look to factor quickly if the numbers are simple. Once factored, set each factor equal to 0 to find all real solutions, and pay close attention to what the question asks for—such as the sum or product of the solutions—rather than stopping after finding just one root.

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Get everything on one side

Try moving the 6 to the left side so that one side of the equation becomes 0. This will give you a standard quadratic equation.

Factor the quadratic

Once you have $x^2 - 5x - 6 = 0$ , look for two numbers that multiply to $-6$ and add up to $-5$ so you can factor the expression.

Use the solutions correctly

After you find the two values of $x$ that satisfy the equation, remember the question is asking for the sum of these distinct real solutions, not just one of them.

Desmos Guide

Enter the equation in Desmos

Type y = x^2 - 5x on one line and y = 6 on another line. This graphs both sides of the original equation.

Find the intersection points

Look for the points where the parabola $y = x^2 - 5x$ intersects the horizontal line $y = 6$ . Note the $x$ -coordinates of these intersection points; these are the real solutions.

Compute the sum of the solutions

Add the $x$ -values of the intersection points (you can use Desmos’s calculator line for this) to get the sum of all distinct real solutions.

Step-by-step Explanation

Rewrite the equation in standard quadratic form

Start with the given equation:

x^2 - 5x = 6

Move 6 to the left side to set the equation equal to 0:

x^2 - 5x - 6 = 0

Now the equation is in standard quadratic form $ax^2 + bx + c = 0$ .

Factor the quadratic and find the solutions

Factor the quadratic expression $x^2 - 5x - 6$ .

We look for two numbers that multiply to $-6$ and add to $-5$ . Those numbers are $-6$ and $1$ :

x^2 - 5x - 6 = (x - 6)(x + 1)

Set each factor equal to 0:

x - 6 = 0 \quad \text{or} \quad x + 1 = 0

So the real solutions are $x = 6$ and $x = -1$ .

Add the distinct real solutions

The distinct real solutions are $6$ and $-1$ .

Add them:

6 + (-1) = 5

So, the sum of all distinct real solutions is $5$ .

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation