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Q: What is the square root of 68 in radical form?

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2 times the square root of 17

Sqrt[5*radical(68)] = -6.4211 and +6.4211.

The idea is to find a perfect square among its factors, such as two square, three square, etc. In this case, the square root of 68, root(68), equals root(4 x 17) = root(4) x root(17) = 2 root(17).

square-root of 68 times square-root of 68 = 68 (square-root of any number) squared is that number

Since 68 is not a perfect square, its square root is irrational.

The square root of 4,624 is 68.

The square root of 68 is 8.246211251 rounded to 9 decimal places.

They are; 8 and 9 but the square root of 68 is about 8.246211251

sqrt(68) = 8.2462 (approx)

The square roots are irrational.

The square roots are irrational.

Rounding, it's 8.25

68

The square root is nearly the inverse operation of squaring or raising to the second power. Nearly but not quite! So sqrt(682) = Â±68

x2+26x = 68 x2+26x-68 = 0 (x+13)2-68 = 0 (x+13)2-169-68 = 0 x+13 = +/- the square root of 237 x = -13 +/- the square root of 237

The square root of 289 is 17 Perimeter = 4*17 = 68 cm

8.2It is: 8.2 rounded to the tenths place

113-2i sqr 17

Yes 8 squared wud equal 64 so that shud be the answer

Don't see any "following" and this I's guessing is what you want? 113-(-68)^.5 = 113-((-1)(68))^.5 = 113-(68)^.5 (-1)^.5 = 113-i(68)^.5

√(68)√(34•2)√(17•2•2)this is as factorable as it gets, notice however, that we have a pair of primes.so we take that pair and pull it out as a single number:2•√(17)√(68)=2•√(17)

pie

68

it is even 8.2 or 8.3

.68