Fun With Roots! Trivia Quiz

Answer: The square root of x is 49

If x is the square root of 7, then the square root of x is actually the fourth root of 7: x = 7^(1/2) implies square root of x = (7^(1/2))^(1/2) = 7^(1/4).

If x is the square root of 7, then by definition, x^2 = 7. The square root of 7 is a real number. (The square root of -7 would be imaginary.) To see that the square root of 7 is irrational, consider the polynomial x^2 - 7. The only possible rational roots of this polynomial are +-1, +-7. None of these are zeros, therefore the square root of 7 is irrational.

Answer: 9

x can be written as 3^(1/2). So x^4 = (3^(1/2))^4 = 3^2 = 9.

Answer: 7

If x is the cube root of 7, then x^3 = 7 by definition.

Answer: 2

There are 4 fourth roots of 16. Let x be a fourth root of 16. Then it satisfies x^4 = 16. Subtracting we get x^4 - 16 = 0. Now the left hand side of the equation can be factored as a difference of squares:

x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4)

So we see the only real fourth roots of 16 are 2 and -2. Setting the other factor to 0 gives us the other two: 2i and -2i (Here i denotes the square root of -1).

Answer: 1

There are 5 fifth roots of 1024, but only one of them is real: 4^5 = 1024 so the (real) fifth root of 1024 is 4.

The other 4 fifth roots of 1024 are all imaginary, but can actually be computed in terms of radicals:

Let a = -1 + sqrt(5),
b = 1 + sqrt(5)
c = sqrt(10 - 2*sqrt(5))
d = sqrt(10 + 2*sqrt(5))

Then the other 4 fifth roots of 1024 are given by:

4e^(2*pi*i/5) = a + di
4e^(4*pi*i/5) = -b + ci
4e^(6*pi*i/5) = -b - ci
4e^(8*pi*i/5) = a - di

Answer: x^2 + 2x + 4 = 0

If x is a cube root of 8, then x satisfies x^3 - 8 = 0. x^3 - 8 factors as a difference of cubes, giving (x - 2)(x^2 + 2x + 4) = 0. Since x is not a real number, x is not 2. Therefore, x^2 + 2x + 4 = 0, and this polynomial doesn't factor any further over the rationals. One way to find the possibilities for x is to use the quadratic formula. We get x = -1 + i*sqrt(3), -1 - i*sqrt(3). We could also use Euler's formula to compute these:

2e^(2*pi*i/3) = 2*(cos(2*pi/3) + i*sin(2*pi/3))
= 2*(-(1/2) + i*sqrt(3)/2)
= -1 + i*sqrt(3)

2e^(4*pi*i/3) = 2*(cos(4*pi/3) + i*sin(4*pi/3))
= 2*(-(1/2) - i*sqrt(3)/2)
= -1 - i*sqrt(3)

Answer: x^2 + 8 = 0

If x is a fourth root of 64, then x satisfies the equation x^4 - 64 = 0. The left hand side factors:
x^4 - 64 = (x^2 - 8)(x^2 + 8)

Setting the first factor to 0 gives the real fourth roots of 64, which are 2*sqrt(2) and -2*sqrt(2). The imaginary fourth roots of 64 come from setting the second factor to 0, giving x^2 + 8 = 0. Solving, we see that the other two fourth roots of 64 are 2*i*sqrt(2), -2*i*sqrt(2).

Answer: x^2 - 4x + 8 = 0, x^2 + 4x + 8 = 0

Let x be a fourth root of -64. Then x satisfies the equation x^4 + 64 = 0. The left hand side can be factored:

x^4 + 64 = (x^4 + 16x^2 + 64) - 16x^2
= (x^2 + 8)^2 - (4x)^2 which is a difference of squares
= (x^2 + 8 - 4x)(x^2 + 8 + 4x)
= (x^2 - 4x + 8)(x^2 + 4x + 8)

Setting these factors to zero and solving give all the fourth roots of -64. Using the quadratic formula, these can be computed quickly:

2 + 2i, 2 - 2i, -2 + 2i, -2 - 2i

Answer: sqrt(6) + sqrt(10)

To see which is largest, square each. Recall that the square of the binomial A+B is given by (A+B)^2 = A^2 + 2AB + B^2.

So 4^2 = 16,
(sqrt{2}+sqrt{14})^2 = 16 + 2*sqrt{28},
(sqrt{5}+sqrt{11})^2 = 16 + 2*sqrt{55},
(sqrt{6}+sqrt{10})^2 = 16 + 2*sqrt{60}.

Since sqrt{6} + sqrt{10} has the largest square, and it is positive, it is the largest of the four numbers.

Answer: sqrt(3)

To see which is the largest, raise each to the sixth power. 1^6 = 1. sqrt(3) to the sixth power is the same as 3^3 = 27. cbrt(5) to the sixth power is the same as 5^2 = 25. Finally, the sixth root of 22 to the sixth power is 22 by definition. Hence sqrt(3) is the largest of the numbers listed.

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