Solving Equations with a Square Root

Squaring can invent an answer, so check each one.

Square both sides to clear the root

Solve √(x + 3) = x − 3. The unknown is under a square root sign. The sign √ always means the square root that is positive or 0, so √9 = 3, never −3.

To remove the root, square both sides. Squaring undoes a square root, so the left side becomes x + 3. The right side becomes (x − 3)², and that bracket must be expanded in full: (x − 3)² = (x − 3)(x − 3) = x² − 6x + 9. So x + 3 = x² − 6x + 9.

Solve the quadratic

Gather every term on one side: subtract x and 3 from both sides to get x² − 7x + 6 = 0. Two numbers that multiply to 6 and add to −7 are −1 and −6, so (x − 1)(x − 6) = 0. The squared equation has two roots, x = 1 and x = 6.

Check each root in the original equation

Put each root back into the equation you started with, √(x + 3) = x − 3, not into the squared one.

For x = 6: the left side is √(6 + 3) = √9 = 3, and the right side is 6 − 3 = 3. The two sides agree, so x = 6 is a solution.

For x = 1: the left side is √(1 + 3) = √4 = 2, but the right side is 1 − 3 = −2. The two sides do not agree, so x = 1 is not a solution. The equation has one solution, x = 6.

Where the extra root came from

Squaring loses the sign of a number. The statement 2 = −2 is false, but squaring both sides gives 4 = 4, which is true. That is exactly what happened at x = 1: the original equation said 2 = −2, and the squared equation said 2² = (−2)².

So the squared equation, x + 3 = (x − 3)², is true in two cases: when √(x + 3) = x − 3, the equation you want, and when √(x + 3) = −(x − 3), which is a different equation. The root x = 1 belongs to the second one. A root that the working produces but the original equation does not have is called an extraneous root.

Because squaring can bring in an extraneous root, the check is not an optional extra here. It is the last step of the method.

xy

The upper curve is y = √(x + 3), and the straight line y = x − 3 meets it only at x = 6. The lower curve is y = −√(x + 3), which squaring lets in: the line meets it at x = 1, the extraneous root.

Get the root alone first

Solve √(2x + 1) + 1 = x. Before squaring, get the square root on its own on one side: subtract 1 from both sides to get √(2x + 1) = x − 1.

Now square both sides: 2x + 1 = x² − 2x + 1. Gather on one side: x² − 4x = 0, so x(x − 4) = 0, and x = 0 or x = 4.

Check in the original equation. For x = 4: √9 + 1 = 3 + 1 = 4, which agrees. For x = 0: √1 + 1 = 2, which is not 0. So x = 0 is extraneous, and the only solution is x = 4.

A quick test before the check

A square root is never negative, so in √(x + 3) = x − 3 the right side must be 0 or more: x − 3 ≥ 0, so x ≥ 3. That rules out x = 1 at a glance. It is a useful warning, but still check every root in the original equation.

The usual mistakes

Squaring (x − 3) as x² + 9 or x² − 9. The bracket is multiplied by itself, (x − 3)(x − 3), which gives the middle term −6x as well: x² − 6x + 9.

Squaring term by term while the root is not alone. √(2x + 1) + 1 = x does not become 2x + 1 + 1 = x², because the square of a sum is not the sum of the squares. Get the root alone first, then square each whole side.

Checking a root in the squared equation. Both roots satisfy it, so that check finds nothing. Check in the equation you started with.

Worked example: The Braking Distance of a Car from Its Whole Stopping Distance, and Whether It Was Speeding

Question On a dry road, a car braking hard slows down by 8 m/s2, so a car that stops over a braking distance of d m was traveling at v = 4√d m/s when the brakes went on. A driver sees a deer on the road. The car keeps its speed for her reaction time of 1 second, and then she brakes to a stop. From the moment she saw the deer to the moment the car stopped, it traveled 60 m. (a) Find the braking distance. (b) The speed limit on the road is 80 km/h. Was the car over the limit when she saw the deer?

  1. 1.Let the braking distance be d m. The car was traveling at 4√d m/s, so in the 1 second before the brakes went on it covered 4√d m. The two distances add up to 60 m: 4√d + d = 60.

    Carreaction 4√dbraking d60 m from seeing the deer to stopping4√d+ d=601 second at 4√dm/s covers 4√dm
    Carreaction 4√dbraking d60 m in all4√d+ d=601 second at 4√dm/s covers 4√dm
    In the 1 second before braking the car covers 4√d m, then it brakes over d m: 4√d + d = 60.

  2. 2.Get the square root alone on one side: 4√d = 60 − d. Square both sides: 16d = 3600 − 120d + d2.

    Carreaction 4√dbraking d60 m from seeing the deer to stopping4√d+ d=604√d=60 − dsubtract d from both sides16d=3600 − 120d + d2square both sides
    Carreaction 4√dbraking d60 m in all4√d+ d=60subtract d from both sides4√d=60 − dsquare both sides16d=3600 − 120d + d2
    Get the square root alone on one side, then square both sides: 16d = 3600 − 120d + d2.

  3. 3.Bring every term to one side: d2 − 136d + 3600 = 0. Two numbers with a product of 3600 and a sum of −136 are −36 and −100, so (d − 36)(d − 100) = 0, and d = 36 or d = 100.

    Carreaction 4√dbraking d60 m from seeing the deer to stopping4√d+ d=604√d=60 − dsubtract d from both sides16d=3600 − 120d + d2square both sidesd2− 136d + 3600=0bring every term to one side(d − 36)(d − 100)=0factorized = 36 or d = 100
    Carreaction 4√dbraking d60 m in all4√d+ d=60subtract d from both sides4√d=60 − dsquare both sides16d=3600 − 120d + d2bring every term to one sided2− 136d + 3600=0factorize(d − 36)(d − 100)=0d = 36 or d = 100
    Bring every term to one side and factorize: (d − 36)(d − 100) = 0.

  4. 4.Squaring can bring in a root that the equation before squaring does not have, so check each one in 4√d = 60 − d. For d = 36: 4 × 6 = 24 and 60 − 36 = 24, which agree. For d = 100: 4 × 10 = 40, but 60 − 100 = −40. Squaring turned 40 = −40 into 1600 = 1600, so d = 100 is rejected; a braking distance of 100 m cannot fit inside a stop of 60 m in any case. (a) The braking distance was 36 m.

    Carreaction 24 mbraking 36 m60 m from seeing the deer to stopping4√d+ d=604√d=60 − dsubtract d from both sides16d=3600 − 120d + d2square both sidesd2− 136d + 3600=0bring every term to one side(d − 36)(d − 100)=0factorized = 36: 4 × 6 = 24, and 60 − 36 = 24d = 100 is rejected: 40 is not −40the braking distance was 36 m
    Carreaction 24 mbraking 36 m60 m in all4√d+ d=60subtract d from both sides4√d=60 − dsquare both sides16d=3600 − 120d + d2bring every term to one sided2− 136d + 3600=0factorize(d − 36)(d − 100)=0d = 36: 4 × 6 = 24, and 60 − 36 = 24d = 100 is rejected: 40 is not −40the braking distance was 36 m
    (a) Only d = 36 satisfies 4√d = 60 − d; d = 100 gives 40 = −40. The braking distance was 36 m.

  5. 5.(b) The speed was v = 4√36 = 24 m/s. An hour has 3600 seconds and a kilometer has 1000 m, so this is 24 × 3.6 = 86.4 km/h. Yes: the car was over the limit of 80 km/h. Check: 24 m during the reaction and 36 m of braking make 60 m.

    Carreaction 24 mbraking 36 m60 m from seeing the deer to stoppingv = 4 ×√36= 24 m/s24 m/s = 24 × 3.6 = 86.4 km/h86.4 km/h is over the limit of 80 km/hcheck: 24 m + 36 m = 60 m
    Carreaction 24 mbraking 36 m60 m in allv = 4 ×√36= 24 m/s24 m/s = 24 × 3.6 = 86.4 km/h86.4 km/h is over the limit of 80 km/hcheck: 24 m + 36 m = 60 m
    (b) The car was traveling at 4√36 = 24 m/s, which is 86.4 km/h: over the limit of 80 km/h.

Answer: (a) 36 m; (b) yes: the car was traveling at 24 m/s, which is 86.4 km/h

Common mistakes

  • Squaring each term of 4√d + d = 60 on its own to get 16d + d2 = 3600. The square of a sum is not the sum of the squares, so get the root alone on one side first and then square each whole side.
  • Keeping d = 100 because it satisfies the quadratic. In the equation before squaring it gives 40 = −40, which is false: squaring made a false equation true, so the root is not a solution.

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