reframing radicals and exponents

An SAT Math micro-topic under Radicals and rational exponents (Advanced Math). Free to read — no account needed.

To simplify a radical, split what is inside into a part that forms complete sets and a leftover part.
Each complete set comes out of the radical as one factor, and whatever is left stays under it.
simplify_radical.png
For a square root, a complete set is a pair, since a2=a\sqrt{a^2} = a.
Take 72\sqrt{72}: since 72=36×272 = 36 \times 2 and 3636 is a perfect square, 72=36×2=62\sqrt{72} = \sqrt{36} \times \sqrt{2} = 6\sqrt{2}.
For a cube root, a complete set is a triple, since a33=a\sqrt[3]{a^3} = a.
Take 243\sqrt[3]{24}: since 24=8×324 = 8 \times 3 and 8=238 = 2^3, 243=83×33=233\sqrt[3]{24} = \sqrt[3]{8} \times \sqrt[3]{3} = 2\sqrt[3]{3}.
Variables work the same way.
For a square root, x5=x4×x=x2x\sqrt{x^5} = \sqrt{x^4} \times \sqrt{x} = x^2\sqrt{x}, since x4x^4 is two complete pairs.
Pull out as many complete sets as you can, and leave the rest inside.

Worked examples

Simplify 50\sqrt{50}.
Split off a perfect square: 50=25×250 = 25 \times 2, and 25=5225 = 5^2.
So 50=25×2=52\sqrt{50} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2}.
Simplify 50x5\sqrt{50x^5}.
Break it into complete pairs and leftovers: 50x5=(25×x4)×(2×x)50x^5 = (25 \times x^4) \times (2 \times x).
The pairs come out as 55 and x2x^2, so 50x5=5x22x\sqrt{50x^5} = 5x^2\sqrt{2x}.
Simplify 24x43\sqrt[3]{24x^4}.
Look for complete triples: 24=23×324 = 2^3 \times 3 and x4=x3×xx^4 = x^3 \times x.
The 232^3 and x3x^3 come out as 22 and xx, leaving 24x43=2x3x3\sqrt[3]{24x^4} = 2x\sqrt[3]{3x}.

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