Equations with radicals — squaring both sides, extraneous solutions: Visual walkthrough

This page rebuilds the whole idea of radical equations as a sequence of pictures. We start from a single line and never use a symbol before you can see it. By the end you will understand — not just remember — why some answers you find are ghosts.

Everything here rests on one small fact about a curve, so we build that curve first.

Step 1 — What a square root actually is (a picture, not a rule)

WHAT. The symbol $x$ is a machine. You feed it a number, it hands you back a number. The rule of the machine: $n$ is the non-negative number whose square is $n$ .

WHY this matters. People say " $9$ is $\pm 3$ ." That is the trap. Both $3$ and $- 3$ square to $9$ , true — but the machine $x$ is only allowed to pick one answer, and it always picks the non-negative one. So $9 = 3$ . Only $3$ . This single restriction is where every phantom answer later comes from.

PICTURE. Look at the blue curve. Its height above each input is $n$ . Notice two things I have marked in red: the curve never dips below the horizontal axis (output is always $\geq 0$ ), and it stops on the left — there is no curve for negative inputs, because no real number squared gives a negative.

Figure — Equations with radicals — squaring both sides, extraneous solutions

Step 2 — The equation we want to solve, drawn as two curves meeting

WHAT. Take a real radical equation: $blue machine’s output x = a straight line x - 6$

WHY draw it this way. An equation "left = right" is a question: for which inputs $x$ do the two sides give the same number? The cleanest way to see that is to draw each side as its own curve and hunt for where they cross. A crossing is a genuine solution; nothing else is.

PICTURE. The blue curve is $x$ (from Step 1). The orange line is $x - 6$ : it rises steadily and passes through height $0$ at $x = 6$ . They cross at exactly one point, near $x = 9$ . Everywhere the orange line sits below zero (all $x < 6$ ), it can never equal the blue curve, because blue is always $\geq 0$ . Circle that region — it is where ghosts will hide.

Step 3 — Squaring: what it does to the two curves

WHAT. To remove the machine, we square both sides. Squaring means "multiply a thing by itself." Applied here: $(x)^{2} = (x - 6)^{2} ⟹ blue got flattened x = orange got folded (x - 6)^{2}$

WHY it changes the game. Squaring $x$ gives back $x$ — clean, the machine is gone. But squaring $x - 6$ folds the whole orange line about the horizontal axis: any part that was below zero flips up to positive. Squaring cannot tell $- 2$ from $+ 2$ ; both become $4$ . That fold is exactly the information we throw away.

PICTURE. On the left, the original orange line dips below zero. On the right (after squaring) the dipped part has been reflected upward into a U-shaped parabola. The blue side became the plain straight line $y = x$ . Watch: the new orange parabola crosses the new blue line in two places now — one is the real crossing from Step 2, the other is brand new, born from the folded-up piece.

Step 4 — Solve the squared equation (find both crossings)

WHAT. Turn $x = (x - 6)^{2}$ into a standard quadratic and factor: $x = x^{2} - 12 x + 36 ⟹ everything on one side x^{2} - 13 x + 36 = 0 ⟹ (x - 4) (x - 9) = 0$

Each bracket, set to zero, gives a candidate: $x = 4$ or $x = 9$ .

WHY factor. A product is zero only when one of its pieces is zero. So $(x - 4) (x - 9) = 0$ splits the problem into two easy questions. These two numbers are exactly the two crossings we saw appear in Step 3's right-hand picture.

PICTURE. The two vertical dashed lines mark $x = 4$ (red) and $x = 9$ (green) landing on the two intersection points of the parabola and the line. Both are honest solutions of the squared equation — but only one of them survived from the original.

Step 5 — The check: send each candidate back to the original two curves

WHAT. Plug each candidate into $x = x - 6$ (the honest Step-2 equation), not the squared one.

$x = 4$ : $4 = 4 - 6 \Rightarrow 2 = - 2$ ✗ $x = 9$ : $9 = 9 - 6 \Rightarrow 3 = 3$ ✓

WHY the check is mandatory. The squared world lost the sign. In the real world the blue machine outputs $+ 2$ at $x = 4$ , but the orange line there is at $- 6 + 4 = - 2$ . They are on opposite sides of zero — no crossing. That is precisely the region we circled in Step 2.

PICTURE. Back on the original two curves: at $x = 9$ blue and orange truly meet (green dot, both at height $3$ ). At $x = 4$ the blue point sits at $+ 2$ while the orange point sits at $- 2$ — I draw the gap between them in red. Same $x$ , different heights: not a solution. That candidate was a reflection artefact from Step 3.

Step 6 — The degenerate case: when every candidate is a ghost

WHAT. Solve $3 x + 4 = - 2$ .

WHY it is special. The whole right side is a negative constant. The blue machine's output can never be negative (Step 1). So blue lives at or above zero, the target line sits flat at $- 2$ below zero — they cannot cross anywhere. No algebra needed: the answer is no solution.

PICTURE. The blue curve floats above the axis; the orange line $y = - 2$ runs flat underneath, forever apart. If you squared anyway you'd get $3 x + 4 = 4$ , so $x = 0$ — but the check gives $4 = 2 \neq = - 2$ . That lone candidate is a pure ghost, manufactured entirely by $(- 2)^{2} = 4$ erasing the minus sign.

Recall Why does a negative right side guarantee no solution?

Because $anything \geq 0$ always, so it can never equal a negative number ::: the two curves live on opposite sides of the horizontal axis and never meet, no matter what $x$ is.

The one-picture summary

Everything above collapses into one story: draw both sides → square (which folds the negative part up) → the fold creates an extra crossing → the check deletes any crossing that lived only in the folded region.

Recall Feynman retelling — say it to a 12-year-old

Imagine two runners on a number-height chart. The blue runner is the square-root machine — she is never allowed below the ground line. The orange runner walks a straight slope and, early on, is underground (negative). We want the moment they are at the exact same height.

Squaring is like flipping the underground part of the orange path up above ground with a mirror. Now the mirrored path touches the blue runner at a spot it never really reached — a fake meeting. When we solve the flipped picture we get two meeting spots, but one of them only happened because of the mirror.

So the last rule is simple and never optional: take every answer back to the real, un-mirrored picture. If both runners are truly at the same height there — keep it. If one is above ground and the other is underground at that same step — throw it out. That thrown-out answer is the extraneous solution.

Practice the reveal

Squaring turns

a = b

into what weaker statement?

a^{2} = b^{2}

, which only forces

a = \pm b

In the picture, an extraneous solution corresponds to a crossing that appears only after which action?

after the fold — squaring reflected the negative part of the line upward

For

f (x) = g (x)

, a candidate is genuine only if

g (x)

is what sign?

non-negative,

g (x) \geq 0

, matching the blue machine's output

Related ideas worth linking in your head: Absolute Value Equations (also about lost signs), Rational Exponents (roots written as powers), and Function Transformations (the fold is a reflection about the axis).