SAT Nonlinear Equations & Systems Question 152 | Free SAT Prep | Aniko

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Question 152·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

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\sqrt{m + 9} = m - 3

What is a solution to the equation above?

Your answer

(Express the answer as an integer)

For radical equations like this, first impose domain restrictions: the expression under a square root must be nonnegative, and the side equal to the root cannot be negative. Then isolate the radical if needed and square both sides once to eliminate it, simplify to a polynomial (often a quadratic), and solve. Always plug each solution back into the original equation to discard any extraneous roots introduced by squaring, which is a common trap on the SAT.

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Think about what values of m are even possible

Because there is a square root on the left and a plain expression $m - 3$ on the right, ask: what must be true about $m + 9$ and $m - 3$ for the equation to make sense?

Remove the square root carefully

Once you know the allowed range for $m$ , try squaring both sides of the equation to eliminate the square root, and then simplify to get a standard algebraic equation.

Watch out for extraneous solutions

After you solve the resulting quadratic, plug each solution back into the original equation (with the square root) to see which ones actually work.

Desmos Guide

Graph both sides of the equation

In Desmos, enter the two expressions as separate functions:

y = sqrt(x + 9)
y = x - 3

Find their intersection point

Zoom or pan until you see where the two graphs cross. Tap on the intersection point; the x-coordinate of this point is the solution to the equation $\sqrt{m + 9} = m - 3$ .

Check the candidate solutions numerically (optional)

You can also type the candidate x-values into Desmos as sqrt(value + 9) and value - 3 to verify that, for the correct solution, both expressions give the same result and the right side is not negative.

Step-by-step Explanation

Set the domain restrictions

The left side is a square root, so the expression inside the root must be nonnegative:

m + 9 \ge 0 \Rightarrow m \ge -9.

The right side is $m - 3$ , and because a square root is never negative, we also need

m - 3 \ge 0 \Rightarrow m \ge 3.

So any solution must satisfy $m \ge 3$ .

Eliminate the square root by squaring

Now square both sides of the equation

\sqrt{m + 9} = m - 3.

Squaring gives

(\sqrt{m + 9})^2 = (m - 3)^2,

m + 9 = (m - 3)^2 = m^2 - 6m + 9.

Move everything to one side to form a quadratic:

0 = m^2 - 6m + 9 - (m + 9) = m^2 - 7m.

Solve the quadratic equation

From

m^2 - 7m = 0,

factor out $m$ :

m(m - 7) = 0.

So the algebraic solutions to the quadratic are

m = 0 \quad \text{or} \quad m = 7.

These are candidate solutions; we still must check them in the original equation and against the restriction $m \ge 3$ .

Check candidate solutions in the original equation

First check $m = 0$ in the original equation $\sqrt{m + 9} = m - 3$ :

Left side: $\sqrt{0 + 9} = \sqrt{9} = 3$ .
Right side: $0 - 3 = -3$ .

They are not equal, and the right side is negative, so $m = 0$ is an extraneous solution.

Now check $m = 7$ :

Left side: $\sqrt{7 + 9} = \sqrt{16} = 4$ .
Right side: $7 - 3 = 4$ .

They match and $m = 7$ satisfies $m \ge 3$ , so the valid solution to the equation is 7.

Strategy

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Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation