# Arithmetic Aptitude :: Square Root and Cube Root

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In the polynomial f(x) =2*x^4 - 49*x^2 +54, what is the product of the roots, and what is the sum of the roots (Note that x^n denotes the x raised to the power n, or x multiplied by itself n times)?

A27,0

B54,2

C49/2,54

D49,27

Explanation:

The given equation is of type ax^4+bx^3+cx^2+dx+e.

Product of roots=(-1)^n*(coeff of constant term/coeff of x^4)= e/a = 54/2 =27

Sum of roots= -(coeff of x^3/coeff of x^4)= -b/a = -0/2=0

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Choose the correct option.

If $$3\sqrt{5} + \sqrt{125}$$ = 17.88, then what will be the value of $$\sqrt{80} + 6\sqrt{5}$$ ?

A13.41

B20.46

C21.66

D22.35

Explanation:

$$3\sqrt{5} + \sqrt{125} = 17.88$$

$$\Rightarrow 3\sqrt{5} + \sqrt{25 \times 5} = 17.88$$

$$\Rightarrow 3\sqrt{5} + 5\sqrt{5} = 17.88$$

$$\Rightarrow 8\sqrt{5} = 17.88$$

$$\Rightarrow \sqrt{5} = 2.235$$

$$\sqrt{80} + 6\sqrt{5} = \sqrt{16 \times 5} + 6\sqrt5$$

= $$4\sqrt5 + 6\sqrt5$$

= $$10\sqrt{5} = (10 \times 2.235) = 22.35$$

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3 / 72

Choose the correct option.

If the product of three consecutive positive integers is 15600 then the sum of the squares of these integers is

A1777

B1785

C1875

D1877

| | | Asked In Green Valley Motor |

Explanation:

Here is no explanation for this answer

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Choose the correct option.

The minimum possible value of the sum of the squares of the roots of the equation x^2 + (a+3)x - (a-5) = 0 is

A1

B2

C3

D4

| | | Asked In Green Valley Motor |

Explanation:

let x and y be the two roots of the equation.

x+y=-(a+3)

x*y=-(a+5)

now, x^2 + y^2 = (x+y)^2 - 2x*y

= a^2 + 9 + 6a + 2a + 10

= a^2 + 8a + 19 = (a+4)^2 +3

the minimum value of this could be when (a+4)^2=0 at a=-4

therefore, minimum value is 3.

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The cube root of .000216 is:

A6

B0.06

C77

D87

Explanation:

$$(.000216)^{\frac{1}{3}} = \left(\frac{216}{10^{6}}\right)^{\frac{1}{3}}$$
= $$\left(\frac{6 \times 6 \times 6 }{10^{2} \times 10^{2} \times 10^{2}}\right)^{\frac{1}{3}}$$
= $$\frac{6}{10^{2}}$$ = $$\frac{6}{100}$$ = 0.06

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What should come in place of both x in the equation $$\frac{x}{\sqrt{128}} = \frac{\sqrt{162}}{X}$$.

A12

B14

C144

D196

Explanation:

Let $$\frac{x}{\sqrt{128}} = \frac{\sqrt{162}}{X}$$
Then$$x^{2} = \sqrt{128 \times 162}$$

=$$\sqrt{64 \times 2 \times 18 \times9}$$

=$$\sqrt{8^2 \times 6^2 \times 3^2}$$

= $$8 \times 6 \times 3$$ = 144.

x =$$\sqrt{144}$$ = 12.

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The least perfect square, which is divisible by each of 21, 36 and 66 is:

A213444

B214344

C214434

D231444

Explanation:

L.C.M. of 21, 36, 66 = 2772.

Now, 2772 = $$2 \times 2 \times 3 \times 3 \times 7 \times 11$$

To make it a perfect square, it must be multiplied by $$7 \times 11$$.

So, required number = $$2^2 \times 3^2 \times 7^2 \times 11^2$$ = 213444

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$$\sqrt{1.5625} = ?$$

A1.05

B1.25

C1.45

D1.55

Explanation:

Here is no explanation for this answer

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If a = 0.1039, then the value of $$\sqrt{4a^2 - 4a + 1} + 3a$$ is:

A0.1039

B0.2078

C1.1039

D2.1039

Explanation:

$$\sqrt{4a2 - 4a + 1} + 3a = \sqrt{(1)^2 + (2a)^2 - 2 x 1 \times 2a }+ 3a$$

= $$\sqrt{(1 - 2a)^2} + 3a$$

= (1 - 2a) + 3a

= (1 + a)

= (1 + 0.1039)

= 1.1039

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If $$x = \frac{\sqrt3 + 1}{\sqrt3-1}$$ and $$y = \frac{\sqrt3 - 1}{\sqrt3+1}$$ , then the value of $$(x^2 + y^2)$$ is:

A10

B13

C14

D15

Explanation:

$$x =\frac{(\sqrt3 + 1)}{(\sqrt3-1)} \times\frac{(\sqrt3 + 1)}{(\sqrt3+1)}= \frac{(\sqrt3 + 1)^2}{3-1} = \frac{3 + 1 + 2\sqrt3}{2} = 2 +\sqrt 3.$$

$$y = \frac{(\sqrt3 - 1)}{(\sqrt3+1)} \times \frac{(\sqrt3 - 1)}{(\sqrt3-1)}= \frac{(\sqrt3 - 1)^2}{3-1} = \frac{3 + 1 - 2\sqrt3}{2} = 2 -\sqrt 3.$$

$$x^2 + y^2 = (2 + \sqrt3)^2 + (2 - \sqrt3)^2$$ = 2(4 + 3) = 14

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