FA4, HA4 - Exact expressions with roots

FA4, HA4 - Exact expressions with roots

A square root can stay exact while you rearrange an expression around it. Learn to collect matching root terms, expand brackets containing roots, and take out common factors. These FA4 and HA4 methods use roots already given in the question, just as earlier algebra used letters.

Like surd terms

One ribbon is 353\sqrt5 metres long and another is 252\sqrt5 metres long. Can you find their exact combined length without a decimal approximation?

Imagine a length of exactly 5\sqrt5 metres as one unit. Three of those units plus two of the same unit make five units:

35+25=553\sqrt5+2\sqrt5=5\sqrt5

The root is a fixed number throughout. You do not need its decimal value to count its multiples.

Exact root terms

A surd is an irrational root left in exact form, such as 2\sqrt2 or 5\sqrt5. Terms such as 353\sqrt5 and −25-2\sqrt5 have the same root factor, so their coefficients can be collected: a5+b5=(a+b)5a\sqrt5+b\sqrt5=(a+b)\sqrt5.

Recall that 5\sqrt5 means the positive square root of 55, not both signs. Its decimal expansion continues without terminating or repeating. A rounded value changes the number; leaving 5\sqrt5 keeps it exact.

CalculationWhat can be combined?Result
43+34\sqrt3+\sqrt3five copies of 3\sqrt3535\sqrt3
43+74\sqrt3+\sqrt7two different root factorskeep both terms
4+34+\sqrt3a whole number and a rootkeep both terms

The coefficient of 3\sqrt3 on its own is 11. A minus sign gives coefficient −1-1.

Worked example 1: collecting a root and constants

Question. Simplify 73+4−23+57\sqrt3+4-2\sqrt3+5.

Plan. Collect the 3\sqrt3-terms together and the constants together.

73+4−23+5=(7−2)3+(4+5)=53+9\begin{aligned} 7\sqrt3+4-2\sqrt3+5 &=(7-2)\sqrt3+(4+5)\\ &=5\sqrt3+9 \end{aligned}

Answer. 53+95\sqrt3+9.

Check. For this check only, use 3≈1.732\sqrt3\approx1.732. The original gives 12.124+4−3.464+5=17.66012.124+4-3.464+5=17.660; the answer gives 8.660+9=17.6608.660+9=17.660. The answer itself stays exact.

Replace the two constants by multiples of a second root. That second group also collects, but the two groups stay separate.

Worked example 2: two distinct root factors

Question. Simplify 73+42−23+527\sqrt3+4\sqrt2-2\sqrt3+5\sqrt2.

Plan. Match identical root factors before adding their signed coefficients.

73+42−23+52=(7−2)3+(4+5)2=53+92\begin{aligned} 7\sqrt3+4\sqrt2-2\sqrt3+5\sqrt2 &=(7-2)\sqrt3+(4+5)\sqrt2\\ &=5\sqrt3+9\sqrt2 \end{aligned}

Answer. 53+925\sqrt3+9\sqrt2.

Check. Rebuild the two groups: 53+23=735\sqrt3+2\sqrt3=7\sqrt3 and 92−52=429\sqrt2-5\sqrt2=4\sqrt2. Both coefficient totals agree with the original groups.

A different sign changes a coefficient, not the root factor.

Watch out

Adding square roots does not mean adding the numbers inside one root. For example, 4+9=2+3=5\sqrt4+\sqrt9=2+3=5, but 4+9=13≠5\sqrt{4+9}=\sqrt{13}\ne5, because 52=255^2=25, not 1313. There is no general rule that replaces a sum of roots by the root of their sum.

The ribbon lengths in the opening question total exactly 555\sqrt5 metres. No rounding is needed.

Expanding exact expressions

Distribution works when a term inside the bracket is an exact root.

Three batches of 2+72+\sqrt7 give three copies of 22 and three copies of 7\sqrt7:

3(2+7)=6+373(2+\sqrt7)=6+3\sqrt7

Treating 7\sqrt7 as one fixed quantity makes this the same step as 3(2+r)=6+3r3(2+r)=6+3r.

Worked example 3: positive distribution

Question. Expand 4(32−5)4(3\sqrt2-5).

Plan. Multiply each bracket term by 44, leaving the root exact.

4(32−5)=4×32+4×(−5)=122−20\begin{aligned} 4(3\sqrt2-5) &=4\times3\sqrt2+4\times(-5)\\ &=12\sqrt2-20 \end{aligned}

Answer. 122−2012\sqrt2-20.

Check. Factor the answer back: both coefficients share 44, giving 4(32−5)4(3\sqrt2-5), the original expression.

Now change only the outside sign. As in ordinary algebra, the negative multiplier applies to both terms.

Worked example 4: negative distribution

Question. Expand −4(32−5)-4(3\sqrt2-5).

Plan. Use −4-4 as the multiplier for both terms.

−4(32−5)=(−4)×32+(−4)×(−5)=−122+20\begin{aligned} -4(3\sqrt2-5) &=(-4)\times3\sqrt2+(-4)\times(-5)\\ &=-12\sqrt2+20 \end{aligned}

Answer. −122+20-12\sqrt2+20.

Check. This expression is the negative of the answer to the previous example, as it should be when the outside multiplier changes from 44 to −4-4. Taking out −4-4 also returns 32−53\sqrt2-5.

If there are root terms outside the bracket, expand before collecting them. For instance,

2(3+5)+5=6+25+5=6+352(3+\sqrt5)+\sqrt5=6+2\sqrt5+\sqrt5=6+3\sqrt5

Each step keeps the fixed root unchanged.

Watch out

424\sqrt2 means 4×24\times\sqrt2. It does not mean 4×2\sqrt{4\times2}. A coefficient outside the root multiplies its value; do not move it inside during expansion.

Expand a fresh bracket with a negative multiplier.

Factoring exact expressions

To take out a common factor, divide every signed term by the chosen factor and put the results in a bracket.

For 6+326+3\sqrt2, the common numerical factor is 33, so the expression is 3(2+2)3(2+\sqrt2). Expanding the product checks both terms. The root itself stays exact.

Worked example 5: a common numerical factor

Question. Take out the greatest common numerical factor from 157+1015\sqrt7+10.

Plan. The coefficients 1515 and 1010 share 55; divide both terms by 55.

157=5(37)10=5(2)157+10=5(37+2)\begin{aligned} 15\sqrt7&=5(3\sqrt7)\\ 10&=5(2)\\ 15\sqrt7+10&=5(3\sqrt7+2) \end{aligned}

Answer. 5(37+2)5(3\sqrt7+2).

Check. Expanding gives 5×37+5×2=157+105\times3\sqrt7+5\times2=15\sqrt7+10.

Change only the sign of the second term. Keep that sign in the bracket.

Worked example 6: a negative term

Question. Take out the greatest common numerical factor from 157−1015\sqrt7-10.

Plan. Again take out 55, this time dividing −10-10 by 55.

157=5(37)−10=5(−2)157−10=5(37−2)\begin{aligned} 15\sqrt7&=5(3\sqrt7)\\ -10&=5(-2)\\ 15\sqrt7-10&=5(3\sqrt7-2) \end{aligned}

Answer. 5(37−2)5(3\sqrt7-2).

Check. The two products are 15715\sqrt7 and −10-10, so expansion restores exactly the original expression.

A given root factor can also be common. In 6x3+936x\sqrt3+9\sqrt3, both terms contain 333\sqrt3:

6x3+93=33(2x+3)6x\sqrt3+9\sqrt3=3\sqrt3(2x+3)

Expanding gives 33×2x+33×33\sqrt3\times2x+3\sqrt3\times3, which restores both terms. This uses distribution with a fixed root constant; no decimal approximation is involved.

Watch out

A factor must be present in every term. In 157−1015\sqrt7-10, the second term has no displayed factor 7\sqrt7. For the requested common numerical factor, use 55; do not replace the bracket's constant by a new root term.

Collecting, expanding and factorising are three ways of reorganising the same exact quantities. That is why the opening ribbon sum can be written as 35+25=(3+2)5=553\sqrt5+2\sqrt5=(3+2)\sqrt5=5\sqrt5.

Quick Recap

  • Keep a given square root as a fixed exact quantity.
  • Collect matching root factors by adding their coefficients; keep distinct groups separate.
  • Distribute a signed multiplier to every term in a bracket.
  • Take out a shared factor and check it by expanding back.

The two ribbons have exact combined length 555\sqrt5 metres.