Two Solutions from One Square

One area, two signs

Watch the completed square become a side, and watch the shared three come away. The equation x squared plus six x minus seven equals zero has a solution of one. A second solution is waiting on the other branch. Why should one square lead to two answers?

Rebuild the reason. Split six x into two strips of three x. Their corner needs nine. This time the constant is negative seven, represented as signed area. Going from negative seven to positive nine requires adding sixteen. Negative area here is algebraic bookkeeping, not a physical patch with a negative size.

Add sixteen to both sides. The completed square has area sixteen and side expression x plus three. We have transformed the original equation into x plus three, all squared, equals sixteen. Each transformation is reversible, so it retains exactly the same solutions.

Take both square roots

Four squared is sixteen. The positive branch therefore gives x plus three equals four. Watch the equal areas become equal side measures. We are choosing a sign for an algebraic quantity. A Euclidean square has a nonnegative side length, but the expression being squared can be negative.

Subtract three from both sides together. The x plus three side loses its three; the four side also loses three. What remains is x equals one. The equal movement matters: subtracting three from just one side would break the equality.

Now choose the minus branch. Negative four also squares to sixteen. So the other equation is x plus three equals negative four. Look at the signed side measure: the negative sign belongs to the root, before we remove the shared three.

Subtract the same three from both sides again. Negative four minus three is negative seven, so the second solution is x equals negative seven. One and negative seven are different solutions of the same original equation. The branch buttons let you compare them without changing its coefficients.

Check both answers in the source equation

See the whole relationship at once: the signed starting areas, the completed square, and the final subtraction. You can also check the original equation by substitution. For x equal to one, one plus six minus seven is zero. For x equal to negative seven, forty-nine minus forty-two minus seven is zero. Both pass.

Consider x squared plus six x plus nine equals zero. Before extracting a side, predict how much is missing. The strips still need a nine-unit corner, and nine is already present. Nothing is missing. The completed square equals zero.

The only real number whose square is zero is zero. Thus x plus three equals zero, and x equals negative three. Plus zero and minus zero coincide, so the two branches give the same root. Two branch buttons do not guarantee two distinct answers.

A second pair of solutions

Try x squared plus four x minus five equals zero. Two strips of width two need a corner of four. Starting at negative five, how much must be added? Predict the completed area and both solutions before moving on. Pause with Interact if you want more time.

The needed addition is nine. The completed side is x plus two, and the positive root of nine is three. Removing two gives one. The other branch uses negative three, and removing two gives negative five. Use the branch buttons yourself to reveal that second result.

Here is the second result: negative five. Return to any example and follow three relationships: the corner fixes the completed area, the square admits both algebraic signs, and equal subtraction isolates x. Those relationships explain the solutions, rather than merely supplying a recipe.

Two roots without integer side lengths

Consider x squared plus four x plus two equals zero. The strip width is two, so the full corner is four. Two units are present and two are missing. Complete the corner by adding two to both sides. We obtain x plus two, all squared, equals two.

The positive square root is square root of two. It is an exact quantity even though its decimal expansion does not terminate. The side expression is x plus two, so removing two gives x equals negative two plus square root of two. Keep the radical until an approximation is useful.

The other branch uses negative square root of two. Removing the same two now gives negative two minus square root of two. The sign choice applies to the square root; it does not reverse the subtraction needed to isolate x. Both branches use the same completed equation.

Return to the positive branch. Square root of two is about one point four one four, so this value of x is about negative zero point five eight six. A positive branch can still produce a negative final answer because we then subtract two. The displayed radical is the exact answer.

The negative branch is about negative three point four one four. Its distance below negative two equals the other root’s distance above negative two. This is a comparison of the displayed algebraic values, not a number-line animation. The square-root magnitude sets the distance and the strip width sets the shift.

Why squaring admits both signs

Return to an exact integer example, x squared plus eight x plus seven equals zero. Completion gives x plus four squared equals nine. At this stage x plus four is the quantity being squared. We first ask which signed values have square nine, before subtracting four to recover x.

The positive choice makes x plus four equal three. Three times three is nine. Removing four gives negative one. A quick substitution into the side expression gives negative one plus four equals three, which returns the chosen positive root before squaring.

The negative choice makes x plus four equal negative three. Negative three times negative three is also nine. Removing four gives negative seven. Substituting into the side expression gives negative seven plus four equals negative three, whose square again returns the same right-hand area.

Both branches therefore satisfy the completed equation. The earlier balanced steps preserve its connection to the original equation. We are not choosing whichever sign gives a preferred-looking answer. A quadratic with this positive completed area has both solutions, and each must be retained unless another stated condition excludes it.

When the two branches meet

Now use x squared plus four x plus four equals zero. The strip width is two and the required corner is four. The corner is already supplied, so completion leaves zero on the right. The equation is x plus two squared equals zero.

The positive square root of zero is zero. Removing two gives negative two. Unlike the earlier example with area two or nine, the square-root distance has disappeared. There is no separation between two values on opposite sides of the shared shift.

Choose the negative branch. Negative zero is still zero, so the final result remains negative two. The two sign choices have produced one distinct solution. It is called a repeated root because the polynomial is the product of the same factor, x plus two, twice.

Reduce the original constant from four to three. The strip width stays two, but one unit is now missing from the corner. Completion gives x plus two squared equals one. Its positive branch is negative one: the root has moved one unit above negative two.

The negative branch is negative three, one unit below negative two. Changing the constant opened the repeated root into two distinct roots while preserving their midpoint. This comparison isolates the effect of the completed area: zero gave no separation; one gives a distance of one on either side.

Compare the positive-area case again: x plus two squared equals two. Its negative branch gives negative two minus square root of two, distinct from the positive branch. Whether the branches are distinct depends on the completed right-hand quantity, not merely on seeing a plus-or-minus symbol in a formula.

Equal distances from the shared shift

Use x squared plus eight x plus seven equals zero. The strips each have width four. Their corner needs sixteen, while seven is already there. Completing leaves nine on the right. Watch the corner become the equal pair of completed areas.

Choose the positive root of nine, which is three. Remove four from both sides. Three minus four is negative one. A positive square-root branch can therefore lead to a negative solution for x; the branch sign and the final answer sign are different questions.

Choose the negative root of nine instead. Negative three minus four gives negative seven. Both answers lie three units from negative four. That midpoint comes from removing the strip width, while the distance comes from the square root of the completed area.

Compare the positive branch once more. Negative one is three above negative four; negative seven was three below it. This is an algebraic comparison of the displayed branch values, not a number line drawn here. Both values return zero when substituted into the original equation.