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# Square Root Questions for RRB Group-D PDF

Download Top-15 RRB Group-D Square root  Questions PDF. RRB GROUP-D Maths questions based on asked questions in previous exam papers very important for the Railway Group-D exam.

Question 1: The square root of $27-4\sqrt{35}$ is :

a) $\pm(\sqrt{5}+2\sqrt{7})$

b) $\pm(\sqrt{5}-2\sqrt{7})$

c) $\pm(\sqrt{7}-2\sqrt{5})$

d) $\pm(\sqrt{7}+2\sqrt{5})$

Question 2: What is the square root of $214 – 78\sqrt{5}$

a) $13-3\sqrt{5}$

b) $17-4\sqrt{5}$

c) $13-6\sqrt{5}$

d) $17-3\sqrt{5}$

Question 3: What is the square root of 97-16$\sqrt{3}$

a) 9-4$\sqrt{3}$

b) 9+4$\sqrt{3}$

c) 7-4$\sqrt{3}$

d) 7+4$\sqrt{3}$

Question 4: Find the square root of discriminant of $2x^2-3x-7 = 0$ ?

a) 65

b) $\sqrt{65}$

c) 56

d) $\sqrt{56}$

Question 5: What can be said about the roots of the following equation $3x^2 – 5x + 4 = 0$

a) They are real and distinct, but one root is not square of the other

b) They are real and distinct, and one root is a perfect square of the other

c) They are real and equal

d) They are complex

Question 6: Sum of the squares of the roots of a quadratic equation is 25 and the product of the roots of the equation is -12. Which of the below four equations can be the quadratic equation –

a) $z^2+3z-12=0$

b) $z^2-z-12=0$

c) $z^2-3z-12=0$

d) $z^2+4z-12=0$

Question 7: If one of the roots of the equation $x^2 + kx – 27 = 0$ is a square of another. What is the value of k?

a) -2

b) -4

c) -6

d) 2

Question 8: The square root of $33-4\sqrt{35}$ is :

a) $\pm(2\sqrt{7}+\sqrt{5})$

b) $\pm(\sqrt{7}+2\sqrt{5})$

c) $\pm(\sqrt{7}-2\sqrt{5})$

d) $\pm(2\sqrt{7}-\sqrt{5})$

Question 9: The greatest 4 digit Number which is a perfect square, is

a) 9999

b) 9909

c) 9801

d) 9081

Question 10: What is the positive square root of $[19+4\sqrt21]?$

a) $\sqrt{7}+2\sqrt{3}$

b) $\sqrt{3}+2\sqrt{7}$

c) $\sqrt{2}+3\sqrt{7}$

d) $\sqrt{7}+3\sqrt{3}$

Question 11: What is the square root of $\frac{(3-2\sqrt2)}{(3+2\sqrt2)}$ ?

a) $3-2\sqrt2$

b) $3+2\sqrt2$

c) $1$

d) $17$

Question 12: What is the square root of $\frac{(3-2\sqrt2)}{(3+2\sqrt2)}?$

a) $3-2\sqrt2$

b) $3+2\sqrt2$

c) $1$

d) 17$Question 13: What is the positive square root of$[25+4\sqrt{39}]$? a)$\sqrt{13}+2\sqrt3$b)$\sqrt{13}+3\sqrt2$c)$\sqrt{11}+2\sqrt3$d)$11+3\sqrt2$Question 14: What is the square root of 156.25 a) 13.5 b) 15.25 c) 10.5 d) 12.5 Question 15: The smallest positive integer to be added to 5467 to make it a perfect square is? a) 9 b) 11 c) 17 d) 25 Answers & Solutions: 1) Answer (C) We have to find$27-4\sqrt{35}$We can write it as : =$\sqrt{{27} – 2 \times 2\times \sqrt{5} \times \sqrt{7}}$Since,$(a^2 + b^2 – 2ab) = (a-b)^2$=$\sqrt{(2\sqrt{5})^2 + (\sqrt{7})^2 – 2\times2\sqrt{5}\times\sqrt{7}}$=$\sqrt{(\sqrt{7} – 2\sqrt{5})^2}$=$\pm(\sqrt{7} – 2\sqrt{5})$2) Answer (A) we have$(a-b)^{2}$=$a^{2}+b^{2}-2ab$Comparing this with$214-78\sqrt{5}$=$a^{2}+b^{2}-2ab$We have 214=$a^{2}+b^{2}$&$ab = 39\sqrt{5}$For a=13 and b=3$\sqrt{5}$; 2ab=2*13*3$\sqrt{5}$And so required answer is 13-3$\sqrt{5}$3) Answer (C) we have$(a-b)^{2}$=$a^{2}+b^{2}-2ab$Comparing this with 97-56$\sqrt{3}$=$a^{2}+b^{2}-2ab$We have 97=$a^{2}+b^{2}$For a=7 and b=4$\sqrt{3}$it gets satisfied and also 2ab=2*7*4$\sqrt{3}$So the$(7-4\sqrt{3})^{2}$=$7^{2}+(4\sqrt{3})^{2}-2*7*4*\sqrt{3}$And so required answer is 7-4$\sqrt{3}$4) Answer (B) Discriminant of$ax^2+bx+c=0$is$b^2-4ac$. Discriminant of$2x^2-3x-7 = 0$is$(-3)^2-4(2)(-7) = 9+56 = 65$Square root of discriminant =$\sqrt{65}$So the answer is option B. 5) Answer (D)$3x^2 – 5x + 4 = 0$Finding discriminant of the given equation we have$(-5)^2 – 4 *3*4$= 25-48 = -23 Since discriminant<0 The roots are complex. Hence, option D is the right answer. 6) Answer (B) Let the roots of the equation be z and y.$ z^2 + y^2 = (z+y)^2 – 2zy$25 = (z+y)^2 +24 z+y = +1 or = -1 So, quadratic equation can be either$z^2+z-12=0$or$z^2-z-12=0$Hence, option B is the right answer. 7) Answer (C) Given equation,$ x^2 + kx – 27 = 0$Let a and$ a^2$be the roots of the equation. We have, a *$a^2$= -27 a = -3 Also, a +$a^2$= -k k = -6 Hence, option C is the right answer. 8) Answer (D) To find :$\sqrt{33-4\sqrt{35}}$We can write it as : =$\sqrt{33 – 2 * 2 * \sqrt{7} * \sqrt{5}}$Since,$(a^2 + b^2 – 2ab) = (a-b)^2$=$\sqrt{(2\sqrt{7})^2 + (5)^2 – 2*2\sqrt{7}*\sqrt{5}}$=$\sqrt{(2\sqrt{7} – \sqrt{5})^2}$=$\pm(2\sqrt{7}-\sqrt{5})$9) Answer (C) Since the number has 4 digits, its square root will always have 2 digits. => Greatest 2 digit no. = 99 Greatest 4 digit no. which is perfect square =$99^2$= 9801 10) Answer (A) Expression :$[19+4\sqrt21]?$=$[19+2\sqrt{4\times21}]=[19+2\sqrt{84}]$=$7+12+2\sqrt{7\times12}$=$(7)^2+(12)^2+2\sqrt{7\times12}$Now, we know that$a^2+b^2+2ab=(a+b)^2$=$(\sqrt{7}+\sqrt{12})^2$Thus, square root is =$\sqrt{7}+\sqrt{12}$=$\sqrt7+2\sqrt3$=> Ans – (A) 11) Answer (A) Given :$x=\frac{(3-2\sqrt2)}{(3+2\sqrt2)}$To find :$\sqrt{x}$Solution : rationalizing the denominator, we get =>$\frac{(3-2\sqrt2)}{(3+2\sqrt2)}\times\frac{(3-2\sqrt2)}{(3-2\sqrt2)}$=$\frac{(3-2\sqrt2)^2}{(3)^2-(2\sqrt2)^2}$=$\frac{(3-2\sqrt2)^2}{9-8}$=>$x=(3-2\sqrt2)^2$Taking square root on both sides, =>$\sqrt{x}=3-2\sqrt2$=> Ans – (A) 12) Answer (A) Expression :$\frac{(3-2\sqrt2)}{(3+2\sqrt2)}$Rationalizing the denominator, =$\frac{(3-2\sqrt2)}{(3+2\sqrt2)}\times\frac{(3-2\sqrt2)}{(3-2\sqrt2)}$=$\frac{(3-2\sqrt2)^2}{(3+2\sqrt2)(3-2\sqrt2)}$=$\frac{(3-2\sqrt2)^2}{9-8}=(3-2\sqrt2)^2$Thus, square root is =$3-2\sqrt2$=> Ans – (A) 13) Answer (A) Expression :$[25+4\sqrt{39}]$=$[25+2\sqrt{4\times39}]=[25+2\sqrt{156}]$=$13+12+2\sqrt{13\times12}$=$(13)^2+(12)^2+2\sqrt{13\times12}$Now, we know that$a^2+b^2+2ab=(a+b)^2$=$(\sqrt{13}+\sqrt{12})^2$Thus, square root is =$\sqrt{13}+\sqrt{12}$=$\sqrt13+2\sqrt3$=> Ans – (A) 14) Answer (D)$15625 = 5*3125 = 25*625 = 25^3 = 5^6$Hence,$\sqrt{15625} = 5^3 = 125$Therefore,$\sqrt{156.25} = 12.5$15) Answer (A) Lets try to find the rough vicinity of the square root of 5467.$70^2 = 4900$and$80^2 = 6400$. Hence, the number is between 70-80.$75^2 = 5625$. Hence, the number is between 70-75 and is likely to be close to 75. We see that$74^2\$ = 5476. Hence, we need to add 9 to make 5467 into a perfect square.

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