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Square Root and Cube Root Questions

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Choose the correct option.

What is the product of the irrational roots of the equation (2x-1)(2x-3)(2x-5)(2x-7)=9?


A3/2

B4

C3

D3/4

Answer: Option A

Explanation:

Let 2x - 4 = y 
=> (y + 3) (y + 1) (y - 1) (y - 3) = 9 
=> (y^2 - 1) (y^2 - 9) = 9 
=> y^4 - 10y^2 + 9 = 9 
=> y^2 (y^2 - 10) = 0 
=> y = 0 or± √(10) 
=> 2x - 4 =± √(10) 
=> x = (1/2) [4± √(10)] 
=> product of the irrational roots 
= (1/2)^2 * [4 + √(10)] [4 - √(10)] 
= (1/4) * 6 
= 3/2.

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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

ENone of these

Answer: Option D

Explanation:

\(3\sqrt{5} + \sqrt{125}\) = 17.88
=> \(3\sqrt{5} + \sqrt{25}*\sqrt{5}\) = 17.88
=> \(3\sqrt{5} + 5\sqrt{5}\) =17.88
=> \(8\sqrt{5}\) = 17.88 => \(\sqrt{5}\) = 2.235

\(\sqrt{80} + 6\sqrt{5}\)
=> \(\sqrt{16} * \sqrt{5} + 6\sqrt{5}\)
=> \(4\sqrt{5} + 6\sqrt{5}\)
=> \(10\sqrt{5)\)

So 10 * 2.235 = 22.35

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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

Answer: Option A

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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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

Answer: Option D

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


A1777

B1785

C1875

D1877

Answer: Option D

Explanation:

Here is no explanation for this answer

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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

Answer: Option C

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

Answer: Option B

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

Answer: Option A

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

Answer: Option A

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

Answer: Option B

Explanation:

Here is no explanation for this answer

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