Question 106·Easy·Nonlinear Equations in One Variable; Systems in Two Variables

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The relationship $z = 3(y - 2)^2$ holds for real numbers $y$ and $z$ . If $y \ge 2$ , which equation correctly expresses $y$ in terms of $z$ ?

A nonlinear formula with a squared group is rearranged by undoing the outside operations first. Divide away the number multiplying the square, then take a square root. Squaring hides whether the group was positive or negative, so the square-root step gives two branches. Use the stated inequality to choose the branch that fits.

Hints

Hint 1
The coefficient is the number multiplying the squared group. Remove that outside factor first: after dividing both sides by $3$ , what remains on the right?
Hint 2
If $u^2=A$ , then $u=\pm\sqrt A$ : a number and its negative have the same square. Here the whole group $y-2$ plays the role of $u$ .
Hint 3
The condition $y\ge2$ tells you whether $y-2$ is negative or nonnegative. Which square-root branch has that sign? Then get $y$ alone.

Step-by-step

Undo the square and choose the sign
Step 1
Get the squared group alone
No Desmos needed. The goal is a formula in $z$ , not a numerical value. In $z=3(y-2)^2$ , the $3$ multiplies the whole squared group. Divide both sides by $3$ :
$\frac{z}{3}=(y-2)^2.$
Step 2
Undo the square
A square root undoes a square, but squaring hides the original sign: a number and its negative have the same square. Keep both signs for now:
$y-2=\pm\sqrt{\frac{z}{3}}.$
Step 3
Use the condition to choose a branch
The given $y\ge2$ means $y-2\ge0$ . The symbol $\sqrt{z/3}$ means the nonnegative square root, so keep the plus branch:
$y-2=\sqrt{\frac{z}{3}}.$
The condition on $y$ decides which sign survives.
Step 4
Get y alone
Add $2$ to both sides:
$y=2+\sqrt{\frac{z}{3}}.$
This expresses $y$ in terms of $z$ while keeping $y\ge2$ . Choice D.

Question 106·Easy·Nonlinear Equations in One Variable; Systems in Two Variables

Strategy

Hints

Step-by-step

Lessons that teach this

Study Hints

Strategy

Hints

Step-by-step

Lessons that teach this