The equation is given. Which choice is the value of ?

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SAT Nonlinear Equations & Systems Question 182 | Free SAT Prep | Aniko

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Question 182·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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The equation $\sqrt{x + 6} + \sqrt{x - 3} = 9$ is given.

Which choice is the value of $x$ ?

For radical equations like this, first determine the domain by requiring the expressions inside square roots to be nonnegative. Then isolate one radical so the equation has the form $\sqrt{\text{expression}} = \text{other expression}$ and square both sides, being very careful to expand any binomials using $(a \pm b)^2 = a^2 \pm 2ab + b^2$ . Simplify to solve for the remaining square root, square again only if necessary, and always substitute your solution back into the original equation to check for extraneous answers created by squaring.

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Think about where square roots are defined

Before solving, make sure the expressions under each square root are nonnegative. What inequality must $x$ satisfy for $x+6$ and $x-3$ to be $\ge 0$ ?

Isolate one radical first

Try moving one of the square root terms to the other side so you have an equation of the form $\sqrt{\text{something}} = \text{expression}$ . That form is easier to square correctly.

Be careful squaring a binomial

When you square $9 - \sqrt{x-3}$ , remember that $(a-b)^2 \neq a^2 - b^2$ . Use $(a-b)^2 = a^2 - 2ab + b^2$ and simplify to isolate the remaining square root.

Don’t forget to check your solution

After solving for $x$ , substitute it back into the original equation with square roots to make sure both sides are equal. This step catches any extraneous (invalid) solutions created by squaring.

Desmos Guide

Enter the two sides of the equation as functions

In one line, type y1 = sqrt(x+6) + sqrt(x-3) and in another line, type y2 = 9. Make sure Desmos is interpreting sqrt correctly (it should show the radical symbol).

Adjust the viewing window and consider the domain

Because both square roots require $x \ge 3$ , set the x-axis to start at about 0 or 2 and extend past 20 or so. Make sure you can see where the curve $y1$ might intersect the horizontal line $y2 = 9$ .

Find the intersection point

Click on the point where the graph of $y1 = \sqrt{x+6} + \sqrt{x-3}$ intersects the line $y2 = 9$ . Desmos will display the coordinates of this point; the x-coordinate of this intersection is the solution to the equation.

Step-by-step Explanation

Set the domain for the square roots

Both square roots must have nonnegative (zero or positive) radicands.

For $\sqrt{x+6}$ , we need $x+6 \ge 0$ , so $x \ge -6$ .
For $\sqrt{x-3}$ , we need $x-3 \ge 0$ , so $x \ge 3$ .

Together, this means any solution must satisfy $x \ge 3$ . We will only consider values of $x$ that are at least 3.

Isolate one radical and square both sides

Start from the equation

\sqrt{x + 6} + \sqrt{x - 3} = 9.

Isolate one square root:

\sqrt{x + 6} = 9 - \sqrt{x - 3}.

Now square both sides. When you square $a-b$ , use $(a-b)^2 = a^2 - 2ab + b^2$ :

\begin{aligned} (\sqrt{x+6})^2 &= (9 - \sqrt{x-3})^2 \\ x + 6 &= 81 - 18\sqrt{x-3} + (x-3). \end{aligned}

Simplify to isolate the remaining square root

Simplify the right-hand side:

Combine the non-radical $x$ terms: $81 + x - 3 = x + 78$ .

So the equation becomes

x + 6 = x + 78 - 18\sqrt{x-3}.

Subtract $x$ from both sides:

6 = 78 - 18\sqrt{x-3}.

Subtract 78 from both sides:

-72 = -18\sqrt{x-3}.

Divide both sides by $-18$ :

\sqrt{x-3} = 4.

Solve for $x$ and check the solution

From $\sqrt{x-3} = 4$ , square both sides:

x - 3 = 16.

Add 3 to both sides:

x = 19.

Check this value in the original equation:

$\sqrt{19+6} = \sqrt{25} = 5$
$\sqrt{19-3} = \sqrt{16} = 4$
$5 + 4 = 9$ , which matches the right side.

Since $x=19$ satisfies the original equation and meets the domain condition $x \ge 3$ , the value of $x$ is $19$ .

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Step-by-step Explanation

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Desmos Guide

Step-by-step Explanation