Raising Equations to Powers: Free Response

Subtract $3$ from both sides to isolate the radical:

$\sqrt{4x + 9} = 7.$

Both sides are now single quantities, so square them to clear the root:

$4x + 9 = 49 \quad\Longrightarrow\quad 4x = 40 \quad\Longrightarrow\quad x = 10.$

Substitute the candidate into the original equation:

$3 + \sqrt{4(10) + 9} = 3 + \sqrt{49} = 3 + 7 = 10.$

The left side matches the right side, so the candidate satisfies the original equation.

Isolating and squaring turned the original equation into a new one, and squaring is an even power. An even power can produce $a = b$ or $a = -b$ from a single squared statement, so the squared equation can be satisfied by a value the original never was:

$a^2 = b^2 \quad\Longrightarrow\quad a = b \ \text{or} \ a = -b.$

That this particular candidate happened to check does not mean skipping the check would have been safe in general; only substitution into the ORIGINAL equation, never the squared one, can tell a genuine candidate from an impostor.

Isolate one radical by moving the other across:

$\sqrt{3x+4} = 2 + \sqrt{x+2}.$

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

$3x + 4 = 4 + 4\sqrt{x+2} + (x+2) = x + 6 + 4\sqrt{x+2}.$

Isolate the surviving radical, then divide by $2$:

$2x - 2 = 4\sqrt{x+2} \quad\Longrightarrow\quad x - 1 = 2\sqrt{x+2}.$

Square a second time and collect into a quadratic:

$(x-1)^2 = 4(x+2) \quad\Longrightarrow\quad x^2 - 6x - 7 = 0 \quad\Longrightarrow\quad (x-7)(x+1) = 0,$

so the candidates are $x = 7$ and $x = -1$.

Substitute each candidate into the ORIGINAL equation, not either squared version reached along the way.

$x=7\!: \ \sqrt{3(7)+4} - \sqrt{7+2} = \sqrt{25} - \sqrt{9} = 5 - 3 = 2 \ \checkmark \qquad x=-1\!: \ \sqrt{3(-1)+4} - \sqrt{-1+2} = \sqrt{1} - \sqrt{1} = 0$

At $x=-1$ the left side is $0$, not $2$, so this candidate fails and is extraneous. The intermediate line $x - 1 = 2\sqrt{x+2}$ already warned us, since its left side must be nonnegative, requiring $x \ge 1$, which $x=-1$ violates.

Squaring is not reversible, because it is an even power: from $a^2=b^2$ you can only conclude $a=b$ or $a=-b$, not $a=b$ alone.

$a^2 = b^2 \quad\Longrightarrow\quad a = b \ \text{or} \ a = -b.$

The first squaring can add a candidate that solves the once-squared equation without solving the original; the second squaring can do the same again, on top of whatever the first one already added. An intermediate equation therefore proves nothing about the original equation. Only substituting directly into the equation as it was first given settles which candidates, if any, are genuine.

Take the square root of both sides, using $\sqrt{(x-5)^2} = \lvert x - 5 \rvert$:

$\lvert x - 5 \rvert = \sqrt{49} = 7 \quad\Longrightarrow\quad x - 5 = \pm 7.$

So $x - 5 = 7$ gives $x = 12$, and $x - 5 = -7$ gives $x = -2$. Both signs are genuine solutions.

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

$3x + 4 = (8-x)^2 = 64 - 16x + x^2.$

Collect into a quadratic and factor:

$x^2 - 19x + 60 = 0 \quad\Longrightarrow\quad (x-4)(x-15) = 0,$

so the candidates are $x=4$ and $x=15$. Test each in the original equation.

For $x=4$: $\sqrt{3(4)+4} = \sqrt{16} = 4$, and $8 - 4 = 4$. They agree.

For $x=15$: $\sqrt{3(15)+4} = \sqrt{49} = 7$, but $8 - 15 = -7$. Since $7 \ne -7$, $x=15$ fails: a principal root can never equal a negative number.

Only substitution into the ORIGINAL equation can tell a genuine candidate from an impostor, because squaring is an even power:

$a^2 = b^2 \quad\Longrightarrow\quad a = b \ \text{or} \ a = -b.$

That is exactly why part B needed a check: the squared equation there admits both branches, and only testing each candidate against the ORIGINAL $\sqrt{3x+4}=8-x$ reveals that $x=15$ actually satisfies the OTHER branch, $\sqrt{3x+4} = -(8-x)$, instead. Part A involved no squaring of the equation at all; it started from an already-squared statement, $(x-5)^2=49$, and undid that square by taking a root. Rooting recovers exactly the values that were squared to produce $49$ in the first place, so it cannot manufacture a value beyond those, and no check against a different equation is needed.

Test each line against the one directly above it.

Line 1 takes the square root of both sides of the given equation: sound, as far as it goes.

Line 2 claims $\sqrt{(x-4)^2} = x - 4$ and $\sqrt{36}=6$, giving $x-4=6$. The right side is fine, but the left side drops the absolute value: $\sqrt{(x-4)^2}$ is $\lvert x - 4 \rvert$, not $x-4$, for every real $x$. The correct line keeps both signs:

$\lvert x - 4 \rvert = 6 \quad\Longrightarrow\quad x - 4 = \pm 6.$

This is the first line that does not validly follow. Lines 3 and 4 carry out valid algebra, but only on the incomplete equation Line 2 handed them, so the negative branch is lost from that point on.

Starting from the corrected line, $x - 4 = \pm 6$, solve each branch separately:

$x - 4 = 6 \ \Rightarrow \ x = 10, \qquad x - 4 = -6 \ \Rightarrow \ x = -2.$

Both values are candidates for $(x-4)^2=36$.

A step that drops a genuine solution, like discarding a sign, makes the final list too short: a value that truly solves the equation is never recovered. A step that fails to check a candidate after an even power, like squaring, makes the final list too long: a value that does not truly solve the equation is kept. In symbols, squaring turns $a=b$ into

$a^2 = b^2 \quad\Longrightarrow\quad a = b \ \text{or} \ a=-b,$

which can add the branch $a=-b$; dropping a sign when undoing that same square instead throws one of the two genuine branches away. The two errors move the solution set in opposite directions, so neither candidate found in part B needs to be checked for being extraneous.

$\sqrt{x^2}$ is defined as the nonnegative number whose square is $x^2$. Since $x \ge 0$ is already nonnegative, and

$x \cdot x = x^2,$

the number $x$ itself satisfies both requirements of that definition. Because the principal square root is unique, $x$ must be that root: $\sqrt{x^2} = x$ when $x \ge 0$.

If $x<0$ then $x$ is negative, and the definition demands a nonnegative output, so $x$ itself is disqualified immediately. Consider $-x$ instead: since $x<0$, $-x>0$, so $-x$ is nonnegative, and

$(-x)^2 = x^2,$

so $-x$ squares back to $x^2$. Both requirements of the definition are satisfied by $-x$, and by uniqueness, $\sqrt{x^2} = -x$ when $x<0$.

Since $\lvert x\rvert = x$ when $x\ge0$ and $\lvert x\rvert=-x$ when $x<0$, parts A and B match it case for case:

$\sqrt{x^2} = x \ (x \ge 0), \qquad \sqrt{x^2} = -x \ (x < 0).$

Together these cover every real $x$, giving $\sqrt{x^2}=\lvert x\rvert$ always. Part A alone only established the nonnegative case; without part B, nothing rules out the identity failing for a negative $x$, where the naive guess $\sqrt{x^2}=x$ is in fact false.