Radical Functions: Free Response

Substitute $h = 150$ into the model and simplify under the root before taking it.

$d(150) = \sqrt{1.5(150)} = \sqrt{225} = 15.$

At $150$ feet the passenger can see $15$ miles to the horizon.

Set $d(h) = 18$. The radical is already isolated, so square both sides to remove it.

$\sqrt{1.5h} = 18 \quad \Longrightarrow \quad 1.5h = 324 \quad \Longrightarrow \quad h = 216.$

Check: $\sqrt{1.5(216)} = \sqrt{324} = 18$, as required.

The square root is an even-index radical, so the radicand cannot be negative. Requiring $1.5h \ge 0$:

$1.5h \ge 0 \quad \Longrightarrow \quad h \ge 0.$

Dividing by the positive $1.5$ keeps the inequality direction unchanged, so the model's domain is every $h \ge 0$. That matches the physical picture directly: a balloon's height above the prairie is never negative, so every height the model actually needs to describe already lies inside the domain the algebra demands.

Square both sides, expanding the right side as a binomial in full:

$x + 3 = (x-3)^2 = x^2 - 6x + 9.$

Collect into a quadratic and factor:

$x^2 - 7x + 6 = 0 \quad \Longrightarrow \quad (x-1)(x-6) = 0,$

so the candidates are $x = 1$ and $x = 6$. Test each in the original equation.

For $x=1$: $\sqrt{1+3} = \sqrt4 = 2$, but $1 - 3 = -2$. Since $2 \ne -2$, $x=1$ fails.

For $x=6$: $\sqrt{6+3} = \sqrt9 = 3$, and $6 - 3 = 3$. They agree, so $x=6$ is genuine.

Square both sides, expanding the right side in full:

$4x + 5 = (x+2)^2 = x^2 + 4x + 4.$

Collect into a quadratic and factor:

$x^2 - 1 = 0 \quad \Longrightarrow \quad (x-1)(x+1) = 0,$

so the candidates are $x=1$ and $x=-1$. Test each in the original equation.

For $x=1$: $\sqrt{4+5} = \sqrt9 = 3$, and $1+2=3$. They agree.

For $x=-1$: $\sqrt{-4+5} = \sqrt1 = 1$, and $-1+2=1$. They agree too.

The first equation, $\sqrt{x+3}=x-3$, shows squaring CAN discard a candidate: $x=1$ solves the squared equation but not the original ($\sqrt4=2\ne1-3=-2$), while $x=6$ is genuine. Taken alone, that might make 'squaring always loses a solution' and 'the smaller one is the loser' look safe.

The second equation, $\sqrt{4x+5}=x+2$, breaks both claims at once. Squaring it produced $x=1$ and $x=-1$, and BOTH check out in the original equation, so no solution was lost at all: squaring does not ALWAYS create an extraneous root. And the surviving pair includes the smaller value, $x=-1$, so being the smaller candidate is no signal of failure either.

What actually decides a candidate's fate is never its size or position; it is only whether substituting it back into the ORIGINAL equation makes that equation true.

$A = B \quad \text{or} \quad A = -B.$

Squaring cannot tell these two branches apart, so each candidate has to be tested on its own terms.

The index is even, so the radicand cannot be negative. Set it $\ge 0$ and solve:

$4x - 20 \ge 0 \quad \Longrightarrow \quad 4x \ge 20 \quad \Longrightarrow \quad x \ge 5.$

Dividing by the positive $4$ keeps the inequality direction, so the domain is every $x \ge 5$.

The index is odd, so the radicand may be positive, negative, or zero; a cube root exists for every real number. For instance,

$(-2)^3 = -8,$

so $\sqrt[3]{-8} = -2$ is a perfectly good real number, and nothing about $x - 6$ can ever make the root undefined. The domain of $g$ is all real numbers.

Require the radicand to be nonnegative:

$9 - 3x \ge 0 \quad \Longrightarrow \quad -3x \ge -9.$

Dividing both sides by $-3$ flips the inequality, since dividing by a negative number reverses which way it points:

$x \le 3.$

The coefficient on $x$ here is $-3$, negative, so isolating $x$ divides by a negative number and flips the inequality. The domain is every $x \le 3$.

Line up the given rule with the general form.

$y = -2\sqrt{x-3}+6 \quad \longleftrightarrow \quad y = a\sqrt{x-h}+k.$

Reading term by term gives $a = -2$, $h = 3$, and $k = 6$. The corner of any curve of this form sits at $(h, k)$, so it is at $(3, 6)$.

The domain comes from the radicand: $x - 3 \ge 0$, so

$x \ge 3.$

For the range, the curve leaves the corner $(3,6)$ heading up when $a>0$ and heading down when $a<0$. Here $a=-2<0$, so the curve descends from the corner and every output is $6$ or less: the range is $y \le 6$.

Shifting a line or a parabola horizontally never removes any input: both accept every real number before and after the shift, because neither one has a radicand to restrict.

A square-root graph is different, because its domain is not a free choice; it is forced by requiring the radicand to be nonnegative. For the unshifted $y=\sqrt{x}$, the radicand is $x$, so the domain is $x \ge 0$. Replacing $x$ with $x-3$ inside the root changes the radicand itself to $x-3$, so the same requirement becomes

$x - 3 \ge 0 \quad \Longrightarrow \quad x \ge 3.$

The corner moving to $x=3$ is not just cosmetic: it marks the new boundary of the domain, because the horizontal shift landed inside the expression the nonnegativity requirement applies to. A line or a parabola has no such expression to shift.

Test each line against the one directly before it.

Line 1 squares both sides of the already-isolated radical equation correctly: sound.

Line 2 claims $(x-5)^2 = x^2+25$. That squares each term of the binomial separately, $x^2$ and $(-5)^2=25$, and simply adds them, dropping the cross term entirely. Expanding the binomial in full,

$(x-5)^2 = (x-5)(x-5) = x^2 - 10x + 25,$

shows Line 2 is missing the $-10x$ term: this is the first line that does not follow.

Lines 3 and 4 carry out valid algebra, but only on the flawed expression Line 2 handed them, so the whole chain from Line 2 onward is unreliable.

Starting from the corrected Line 2, $49-5x = x^2-10x+25$, collect every term onto one side:

$0 = x^2 - 10x + 25 - 49 + 5x = x^2 - 5x - 24.$

Factor the quadratic:

$x^2 - 5x - 24 = (x-8)(x+3) = 0,$

so the candidates are $x = 8$ and $x = -3$.

Substitute each candidate into the ORIGINAL equation, never the squared one.

For $x = 8$:

$\sqrt{49-5(8)} = \sqrt9 = 3, \qquad 8 - 5 = 3.$

The two sides agree, so $x=8$ is genuine.

For $x = -3$:

$\sqrt{49-5(-3)} = \sqrt{64} = 8, \qquad -3 - 5 = -8.$

Since $8 \ne -8$, $x=-3$ fails the original equation and is extraneous: the square root can only equal its principal, nonnegative value $8$, never $-8$. Correcting the algebra did not remove the need to check every candidate in the original equation; only substitution decides which survive.