Q: In the following determine the set of value of P for which the given quadratic equation has real roots. Px2 + 4 x + 1 = 0

Given :- Px2 + 4x + 1 = 0 Compare with Ax 2 + Bx + C = 0, we get A = P, B = 4, C = 1 For real roots, B2 - 4AC ≥ 0 ⇒ 16 - 4P ≥ 0 ⇒ 16 ≥ 4P ⇒ P ≤ 16/4 ⇒ P ≤ 4. Hence , required answer is option C .

Q: With respect to the roots of x2 - x - 2 = 0, we can say that :

Given :- x2 - x - 2 = 0 The given equation is of the form ax2 + bx + c = 0 Here , a = 1 , b = -1 , c = - 2 Now, D2 = b2 - 4ac = ( -1 )2 - 4 x 1 x ( - 2 ) D = √9 = 3 So, roots are rational . Hence, both the roots must be integers.

Question 1

In solving a problem, one student makes a mistake in the coefficient of the first degree term and obtain -9 and -1 for the roots. Another student makes a mistake in the constant term of the equation and obtains 8 and 2 for the roots. The correct equ…

Accepted Answer

When mistake is done in first degree term, the roots of the equation are -9 and -1. ∴ Equation (x+ 1) (x + 9) = x2 + 10x + 9 ...(i) When mistake is done in constant term, the roots of equation are 8 and 2. ∴ Equation is (x - 2) (x - 8) = x2 - 10x + 16 .....(ii) ∴ Required equation from Eqs. (i) and (ii) is = x2 - 10x + 9 Also we see in both the cases 1st degree term is same with oposite sign i.e., in such questions we should take data from given conditions and find the correct equation.

Question 2

If one of the roots of the equation x 2 - bx + c = 0 is the square of the other, then which of the following option is correct?

Accepted Answer

Given that, one root of the equation x2 - bx + c = 0 is square of other root of this equation i.e., roots (α, α2). ∴ Sum of roots = α + α2 = -(-b)/1 ⇒ α (α + 1) = b .......(i) and product of roots= α. α2 = c/1 ⇒ α3 = c ⇒ = c1/3 ....(ii) From Eqs. (i) and (ii). c1/3 (c1/3 + 1) = b ....(iii) On cubing both sides, we get c(c1/3 + 1)3 = b3 ⇒ c {c + 1 + 3c1/3 (c1/3 + 1)} = b3 ⇒ c {c + 1 = 3b} = b3 [from Eq. (iii)] ⇒ b3 = 3bc + c2 + c

Question 3

Two students A and B solve an equation of the from x 2 + px + q = 0. A starts with a wrong value of p and obtains the roots as 2 and 6. B starts with a wrong value of q and gets the roots as 2 and -9. What are the correct roots of the equation?

Accepted Answer

Let α and β be the roots of the quadratic equation x2 + px + q = 0 Given that, A starts with a wrong value of p and obtains the roots as 2 and 6. But this time q is correct. i.e products of roots = q = α . β = 6 x 2 = 12 ...(i) and B starts with a wrong value of q and gets the roots as 2 and - 9. But this time p is correct i.e., sum of roots = p = α + β = - 9 + 2 = - 7 ..(ii) (α - β)2 = (α + β)2 - 4α β = (-7)2 - 4.12 = 49 - 48 = 1 [from Eqs. (i) and (ii)] ⇒ α - β = 1 ..(iii) Now, from Eqs. (ii) and (iii) , we get α = - 3 and β = - 4 which are correct roots .

Question 4

If x = √ √(5 + 1)/√(5 - 1), then x 2 - x - 1 is equal to

Accepted Answer

Here, x = √√5 + 1/√5 - 1 On rationalising the terms given in square root, we get x = √√5 + 1/√5 - 1 x √5 + 1/√5 + 1 √5 + 1/2 Now, substituting the value of x in x2 - x - 1. ∴ x2 - x - 1 = (√5 + 1/2)2 - (√5 + 1/2) - 1 = 5 + 1 + 2√5/4 - √5 + 1/2 - 1 = 6 + 2√5 - 2√5 - 2 - 4 /4 = 0

Question 5

If α, β are the roots of the equation x2 - 5x + 6 = 0, construct a quadratic equation whose roots are 1 , 1 . α β

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Question 6

Consider the equation px2 + qx + r = 0, where p, q, r are real. The roots are equal in magnitude but opposite in sign when :

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Question 7

The roots of the equation 4x - 3 × 2x × 22 + 32 = 0 would include :

Accepted Answer

According to question , we have Given equation is :- 22x - 3 × 2x × 22 + 32 = 0 ⇒ 22x - 12 × 2x + 32 = 0 ⇒ y2 - 12y + 32 = 0, where 2x = y ⇒ ( y - 8 ) ( y - 4 ) = 0 ⇒ y = 8, y = 4 ∴ 2x = 8 or, 2x = 4 ⇒ 2x = 23 or, 2x = 22 ⇒ x = 3 or, x = 2. Therefore , the roots of the equation are 3 and 2.

Question 8

Find the value of k so that the sum of the roots of the equation 3x2 + (2x + 1)x - k - 5 = 0 is equal to the product of the roots :

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Question 9

Ⅰ. x 2 -24x + 144 = 0 Ⅱ. y 2 − 26y + 169 = 0

Accepted Answer

As per the given above question , we have From equation Ⅰ. x2 - 24x + 144 = 0 ⇒ x2 − 12x - 12x + 144 = 0 ⇒ x(x − 12) − 12(x − 12) = 0 ⇒ (x − 12)2 = 0 ∴ x = 12 From equation Ⅱ. y2 − 26y + 169 = 0 ⇒ y2 − 13y − 13y + 169 = 0 ⇒ y(y − 13) − 13(y − 13) = 0 ⇒ (y − 13)2 = 0 ∴ y = 13 Hence, required answer will be x < y .

Question 10

Ⅰ. 2x 2 + 3x – 20 = 0 Ⅱ. 2y 2 + 19y + 44 = 0

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Question 11

Ⅰ. 10x 2 − 7x + 1 = 0 Ⅱ. 35y 2 -12y + 1 = 0

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Question 12

Ⅰ. 6x 2 + 77x + 121 = 0 Ⅱ. y 2 + 9y − 22 = 0

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Question 13

Ⅰ. x 2 − 6x = 7 Ⅱ. 2y 2 + 13y + 15 = 0

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Question 14

In the following determine the set of value of P for which the given quadratic equation has real roots. Px2 + 4 x + 1 = 0

Accepted Answer

Given :- Px2 + 4x + 1 = 0 Compare with Ax 2 + Bx + C = 0, we get A = P, B = 4, C = 1 For real roots, B2 - 4AC ≥ 0 ⇒ 16 - 4P ≥ 0 ⇒ 16 ≥ 4P ⇒ P ≤ 16/4 ⇒ P ≤ 4. Hence , required answer is option C .

Question 15

The roots of the equation √7x2 - 6x - 13√7 = 0 are

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Question 16

In the following, find the value (s) of P so that the given equation has equal roots 3x 2 - 5x + P = 0

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Question 17

With respect to the roots of x2 - x - 2 = 0, we can say that :

Accepted Answer

Given :- x2 - x - 2 = 0 The given equation is of the form ax2 + bx + c = 0 Here , a = 1 , b = -1 , c = - 2 Now, D2 = b2 - 4ac = ( -1 )2 - 4 x 1 x ( - 2 ) D = √9 = 3 So, roots are rational . Hence, both the roots must be integers.

Question 18

If α, β are the roots of the quadratic equation x2 - 8x + k = 0, find the value of k such the α2 + β2 = 40

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Question 19

Ⅰ. x 2 – 11x + 24 = 0 Ⅱ. 2y 2 – 9y + 9 = 0

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Question 20

Ⅰ. 2 × 4 - 3 × 4 = x10/7 x 4/7 x 4/7 Ⅱ. y 3 + 783 = 999

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Quadratic Equation Questions