Q: x 2 - 365 = 364; y - √ 324 = √ 81

x2 - 365 = 364 ⇒ x2 = 364 + 365 ∴ x = √729 = ± 27 and y - √324 = √81 ⇒ y - 18 = 9 ∴ y = 27 So, y ≥ x or x ≤ y because y = 27 and x = 27 and - 27.

Q: (4/√ x) + (7/√ x) = √ x; y 2 - (11) 5/2/√ y = 0

4/√x + 7/√x = √x; ⇒ 11/√x = √x ∴ x = 11 and y2 - (11)5/2/√y = 0 ⇒ y2 = (11)5/2/(y)1/2 ⇒ y2 x y1/2 = (11)5/2 ⇒ (y)5/2 = (11)5/2 ∴ y = 11 ∴ x = y

Q: Mr. Arjun is on tour and he has ₹ 360 for his expenses. If he exceeds his tour by 4 days, he must cut down his daily expenses by ₹ 3. For how many days is Mr. Arjun out on tour?

Let Mr . Arjun is on tour for N days. Then, according to the question, Difference in expenses per day for original and extended tour = 3 360/N - 360/(N + 4) = 3 ⇒ 1/N - 1/(N + 4) = 3/360 = 1/120 ⇒ [N + 4 - N] / [ N x (N + 4)] = 1/120 ⇒ N(N + 4) = 4 x 120 = 480 ⇒ N2 + 4N - 480 = 0 ⇒ N2 + 24 - 20x - 480 = 0 ⇒ N(N + 24) - 20(N + 24) = 0 ⇒ (N + 24) (N - 20) = 0 ∴ N = 20 (we ignore -ve value of x, that is, -24 because days cannot be negative)

Q: The quadratic equation with rational coefficients, whose one root is √5, is :

As per the given above question , we have Let k1 and k2 be the roots of quadratic equation . One root is k1 = √5 So the other root is k2 = − √5 ∴ Sum of the roots = k1 + k2 = √5 + ( -√5 ) = 0 and product of the roots = k1 × k2 = − (√5) (- √5) = −5 ∴ Required equation is x2 - (sum of the roots)x + (product of the roots) = 0 ⇒ x2 - 5 = 0.

Q: Ⅰ. x 2 – 19x + 84 = 0 Ⅱ. y 2 – 25y + 156 = 0

According to question ,we can say that From equation Ⅰ. x2 – 19x + 84 = 0 ⇒ x2 – 7x – 12x + 84 = 0 ⇒ (x – 7)(x – 12) = 0 ⇒ x = 7, 12 From equation Ⅱ. y2 – 25y + 156 = 0 ⇒ y2 – 13y – 12y + 156 = 0 ⇒ (y – 13)(y – 12) = 0 ⇒ y =12, 13 ∴ x ≤ y From above it is clear that x ≤ y is correct answer .

Q: Ⅰ. √ 784x + 1234 = 1486 Ⅱ. √ 1089y + 2081 = 2345

According to question ,we can say that From equation Ⅰ. √784x + 1234 = 1486 ⇒ √784x = 252 ⇒ 28x = 252 ⇒ x = 9 From equation Ⅱ. √1089y + 2081 = 2345 ⇒ √1089y = 264 ⇒ 33y = 264 ⇒ y = 8 ∴ x > y Thus , required answer will be x > y .

Question 1

x 2 - 8x + 15 = 0 ; y 2 - 3y + 2 = 0

Accepted Answer

x2 - 8x + 15 = 0 ⇒ x2 - 5x - 3x + 15 = 0 ⇒ x(x - 5) -3(x - 5) = 0 ⇒ (x - 5) (x - 3) = 0 ∴ x = 5, 3 and y2 - 3y + 2 = 0 ⇒ y2 - 2y - y + 2 = 0 ⇒ y(y - 2) -1(y - 2) = 0 ⇒ (y -2) (y - 1) = 0 ⇒ y = 2, 1 ∴ x > y

Question 2

x 2 - x - 12 = 0; y 2 + 5y + 6 = 0

Accepted Answer

x2 - x - 12 = 0 ⇒ x2 - 4x + 3x - 12 = 0 ⇒ x(x - 4) + 3 (x - 4) = 0 ⇒ (x - 4) (x + 3) = 0 ∴ x = - 3, 4 and y2 + 5y + 6 = 0 ⇒ y2 + 3y + 2y + 6 = 0 ⇒ y(y + 3) + 2(y + 3) = 0 ⇒ (y + 3) (y + 2) = 0 ⇒ y = - 3, - 2 ∴ x ≥ y [∵ x = - 3 and y = - 3, so x = y and x = 4 and y = - 2, hence x > y]

Question 3

3x 2 + 8x + 4 = 0; 4y 2 - 19y + 12 = 0

Accepted Answer

3x2 + 8x + 4 = 0 ⇒ 3x2 + 6x + 2x + 4 = 0 ⇒ 3x(x + 2) + 2(x + 2) = 0 ⇒ (x + 2) (3x + 2) = 0 ∴ x = - 2, - 2/3 and 4y2 - 19y + 12 = 0 ⇒ 4y2 - 16y - 3y + 12 = 0 ⇒ 4y(y - 4) -3(y - 4) = 0 ⇒ (y - 4) (4y - 3) = 0 ∴ y = 4, 3/4 Hence, y > x or x < y

Question 4

x 2 - 365 = 364; y - √ 324 = √ 81

Accepted Answer

x2 - 365 = 364 ⇒ x2 = 364 + 365 ∴ x = √729 = ± 27 and y - √324 = √81 ⇒ y - 18 = 9 ∴ y = 27 So, y ≥ x or x ≤ y because y = 27 and x = 27 and - 27.

Question 5

(4/√ x) + (7/√ x) = √ x; y 2 - (11) 5/2/√ y = 0

Accepted Answer

4/√x + 7/√x = √x; ⇒ 11/√x = √x ∴ x = 11 and y2 - (11)5/2/√y = 0 ⇒ y2 = (11)5/2/(y)1/2 ⇒ y2 x y1/2 = (11)5/2 ⇒ (y)5/2 = (11)5/2 ∴ y = 11 ∴ x = y

Question 6

Mr. Arjun is on tour and he has ₹ 360 for his expenses. If he exceeds his tour by 4 days, he must cut down his daily expenses by ₹ 3. For how many days is Mr. Arjun out on tour?

Accepted Answer

Let Mr . Arjun is on tour for N days. Then, according to the question, Difference in expenses per day for original and extended tour = 3 360/N - 360/(N + 4) = 3 ⇒ 1/N - 1/(N + 4) = 3/360 = 1/120 ⇒ [N + 4 - N] / [ N x (N + 4)] = 1/120 ⇒ N(N + 4) = 4 x 120 = 480 ⇒ N2 + 4N - 480 = 0 ⇒ N2 + 24 - 20x - 480 = 0 ⇒ N(N + 24) - 20(N + 24) = 0 ⇒ (N + 24) (N - 20) = 0 ∴ N = 20 (we ignore -ve value of x, that is, -24 because days cannot be negative)

Question 7

The quadratic equation with rational coefficients, whose one root is √5, is :

Accepted Answer

As per the given above question , we have Let k1 and k2 be the roots of quadratic equation . One root is k1 = √5 So the other root is k2 = − √5 ∴ Sum of the roots = k1 + k2 = √5 + ( -√5 ) = 0 and product of the roots = k1 × k2 = − (√5) (- √5) = −5 ∴ Required equation is x2 - (sum of the roots)x + (product of the roots) = 0 ⇒ x2 - 5 = 0.

Question 8

The roots of x + 4 + x - 4 = 10 are : x - 4 x + 4 3

Accepted Answer

Question 9

Ⅰ. x 2 – 19x + 84 = 0 Ⅱ. y 2 – 25y + 156 = 0

Accepted Answer

According to question ,we can say that From equation Ⅰ. x2 – 19x + 84 = 0 ⇒ x2 – 7x – 12x + 84 = 0 ⇒ (x – 7)(x – 12) = 0 ⇒ x = 7, 12 From equation Ⅱ. y2 – 25y + 156 = 0 ⇒ y2 – 13y – 12y + 156 = 0 ⇒ (y – 13)(y – 12) = 0 ⇒ y =12, 13 ∴ x ≤ y From above it is clear that x ≤ y is correct answer .

Question 10

Ⅰ. 12 - 23 = 5√x √ x √ x Ⅱ. √ y - 5 √ y = 1 12 12 √ y

Accepted Answer

Question 11

Ⅰ. 9 + 19 = √x √ x √ x Ⅱ. y5 - ( 2 × 14 ) 11/2 = 0 √ y

Accepted Answer

Question 12

Ⅰ. √ 784x + 1234 = 1486 Ⅱ. √ 1089y + 2081 = 2345

Accepted Answer

According to question ,we can say that From equation Ⅰ. √784x + 1234 = 1486 ⇒ √784x = 252 ⇒ 28x = 252 ⇒ x = 9 From equation Ⅱ. √1089y + 2081 = 2345 ⇒ √1089y = 264 ⇒ 33y = 264 ⇒ y = 8 ∴ x > y Thus , required answer will be x > y .

Question 13

Ⅰ. x 3 – 468 = 1729 Ⅱ. y 2 – 1733 + 1564 = 0

Accepted Answer

According to question, From equation Ⅰ. x3 – 468 = 1729 ⇒ x3 = 2197 = 133 ⇒ x = 13 From equation Ⅱ. y2 – 1733 + 1564 = 0 ⇒ y2 = 169 = 132 ⇒ y = ±13 Thus , x ≥ y is required answer .

Question 14

Ⅰ. x 2 – 7x + 12 = 0 Ⅱ. y 2 + y – 12 = 0

Accepted Answer

From the given equations , we can say that From equation Ⅰ. x2 - 7x + 12 = 0 ⇒ x2 - 4x - 3x + 12 = 0 ⇒ x( x - 4 ) - 3( x - 4 ) = 0 ⇒ ( x - 3 )( x - 4 ) = 0 ⇒ x = 3 or x = 4 From equation Ⅱ. y2 + y - 12 = 0 ⇒ y2 + 4y - 3y - 12 = 0 ⇒ y( y + 4 ) - 3( y + 4 ) = 0 ⇒ ( y - 3 )( y + 4 ) = 0 ⇒ y = 3 or y = - 4 From above both equations it is clear that x ≥ y is correct answer .

Question 15

Ⅰ. x 4 – 227 = 398 Ⅱ. y 2 + 321 = 346

Accepted Answer

According to question ,we can say that From equation Ⅰ. x4 = 398 + 227 = 625 ⇒ x4 = 54 From equation Ⅱ. y2 = 346 - 321 = 25 ⇒ y2 = 52

Question 16

Ⅰ. x 2 - 1 = 0 Ⅱ. y 2 + 4y + 3 = 0

Accepted Answer

From the given equations , we know that Ⅰ. x2 - 1 = 0 ⇒ x2 = 1 ⇒ x = ± √1 = ± 1 Ⅱ. y2 + 4y + 3 = 0 ⇒ y2 + y + 3y + 3 = 0 ⇒ y( y + 1) + 3( y + 1) = 0 ⇒ ( y + 1) ( y + 3) ⇒ y = -3 or -1 From above equations it is clear that x ≥ y is correct answer .

Question 17

Ⅰ. x 3 - 371 = 629 Ⅱ. y 3 - 543 = 788

Accepted Answer

According to given question , we have Ⅰ. x3 - 371 = 629 ⇒ x = √1000 = 10 Ⅱ. y3 - 543 = 788 ⇒ y = √1331 = 11 From above both equations it is clear that x < y is correct answer .

Question 18

Ⅰ. 2x 2 + 11x + 12 = 0 Ⅱ. 5y 2 + 27y + 10 = 0

Accepted Answer

Question 19

If one root of the quadratic equation 2x 2 + Px + 4 = 0 is 2, find the second root and value of P.

Accepted Answer

Question 20

The roots of the equation 3a2x2 - abx - 2b2 = 0 are

Accepted Answer

Quadratic Equation Questions