Q: x = 3a solution of the equation 3x2 + (k − 1)x + 9 = 0 if k has value

The given quadratic equation 3x2 + (k − 1)x + 9 = 0 Putting x = 3 in above given eq. , we get 27 + 3(k − 1) + 9 = 0 ⇒ 27 + 3k - 3 + 9 = 0 ⇒ 3k = −33 ⇒ k = −11. Hence , the value of k is −11.

Q: The expression x4 + 7x2 + 16 can be factored as :

The given expression x4 + 7x2 + 16 = ( x4 + 8x2 + 16 ) - x2 = ( x4 + 2 * 4 * x2 + 42 ) - x2 = ( x2 + 4 )2 - x2 = ( x2 + 4 + x ) ( x2 + 4 - x ) Hence , x4 + 7x2 + 16 = ( x2 + x + 4 ) ( x2 - x + 4 ). Option C is correct answer .

Question 1

If α, β are the roots of the equation 2x2 - 3x + 1 = 0, form an equation whose roots are α and β β α

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Question 2

The factors of ( a2 + 4b2 + 4b – 4ab – 2a – 8 ) are :

Accepted Answer

Given that :- The factors of a2 + 4b2 + 4b – 4ab – 2a – 8 = a2 + 4b2 – 4ab – 2a + 4b – 8 = (a – 2b) 2 – 2(a – 2b) – 8 Let, (a – 2b) = x ∴ The given expression = x2 – 2x – 8 = x2 – 4x + 2x – 8 = x(x – 4) + 2(x – 4) = (x – 4) (x + 2) Putting the value of x , we get ∴ a2 + 4b2 + 4b – 4ab – 2a – 8 = (a – 2b – 4) (a – 2b + 2) Therefore , required answer will be (a – 2b – 4) (a – 2b + 2) .

Question 3

Find the value of √30 + √30 + √30 +........

Accepted Answer

According to question ,we can say that Let , x = √30 + √30 + √30 +........ On squaring both sides, we have x2 = 30 + √30 + √30 + √30 +........ ⇒ x2 = 30 + x ⇔ x2 - x - 30 = 0 ⇒ x2 - 6x + 5x - 30 = 0 ⇒ x( x - 6 ) + 5( x - 6 ) = 0 ⇒ ( x - 6 ) ( x + 5 ) = 0 ⇒ x = 6 because x ≠ - 5 Hence , required answer is 6 .

Question 4

The sum of the squares of 2 natural consecutive odd numbers is 394. The sum of the numbers is :

Accepted Answer

Let, the two natural consecutive odd numbers be n and (n + 2) Now, according to the question, ⇒ n2 + ( n + 2 )2 = 394 ⇒ n2 + n2 + 4 + 4n = 394 ⇒ 2n2 + 4n - 390= 0 ⇒ n2 + 2n - 195 = 0 ⇒ n2 + 15n - 13n - 195 = 0 ⇒ n( n + 15 ) - 13( n + 15 ) = 0 ⇒ ( n + 15 ) ( n - 13 ) = 0 ⇒ n = 13 and n ≠ - 15 ∴ The two natural consecutive odd numbers be 13 and (13 + 2) = 15 . ∴ the sum of the numbers = 13 + 15 = 28 Quicker Approach: By mental operation, 132 + 152 = 169 + 225 = 394 ∴ Required sum = 13 + 15 = 28

Question 5

The roots of the equation ax2 + bx + c = 0 will be reciprocal if

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Question 6

Ⅰ. x 2 – 4 = 0 Ⅱ. y 2 + 6y + 9 = 0

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According to question ,we have From equation Ⅰ. x2 – 4 = 0 ⇒ x = ± 2, From equation Ⅱ. y2 + 6y + 9 = 0 ⇒ ( y + 3 )2 = 0 ⇒ ( y + 3 )( y + 3 ) = 0 ⇒ y + 3 = 0 ⇒ y = -3 , -3 Clearly , x > y is required answer .

Question 7

Ⅰ. x 2 = 729 Ⅱ. y = √ 729

Accepted Answer

According to question ,we have From equation Ⅰ. x2 = 729 ⇒ x = ± √729 = ± 27 From equation Ⅱ. y = √729 ⇒ y = 27 Hence , x ≤ y is required answer .

Question 8

Ⅰ. 2x 2 + 11x + 14 = 0 Ⅱ. 4y 2 + 12y + 9 = 0

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Question 9

Ⅰ. x 2 - 7x + 12 = 0 Ⅱ. y 2 - 12y + 32 = 0

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As per the given above equations , we have From equation Ⅰ. x2 - 7x + 12 = 0 ⇒ x2 - 4x - 3x + 12 = 0 ⇒ x( x - 4) - 3( x - 4) = 0 ⇒ ( x - 4) ( x - 3) = 0 ⇒ x = 4 or, 3 Ⅱ. y2 - 12y + 32 = 0 ⇒ y2 - 8y - 4y + 32 = 0 ⇒ y( y - 8) - 4( y - 8) = 0 ⇒ ( y - 8) ( y - 4) = 0 ⇒ y = 4 or, 8 Thus , x ≤ y is required answer .

Question 10

Ⅰ. 5x + 2y = 31 Ⅱ. 3x + 7y = 36

Accepted Answer

Given that :- Given that :- Ⅰ. 5x + 2y = 31 .....( 1 ) Ⅱ. 3x + 7y = 36 .....( 2 ) Solving these two linear equations, we get x = 5, y = 3 . ∴ x > y is correct answer .

Question 11

The equation x2 − px + q = 0, p, q ∈ R has on real root if :

Accepted Answer

According to question , we can say that The equation x2 − px + q = 0, p, q ∈ R has no real root On comparing with quadratic eq. Ax2 + Bx + C = 0 , we get ∴ A =1, B= - p, C = q Now ,we have B2 < 4AC ⇒ ( - P )2 < 4 x 1 x q ⇒ p2 < 4q. Hence , option B is correct answer .

Question 12

x = 3a solution of the equation 3x2 + (k − 1)x + 9 = 0 if k has value

Accepted Answer

The given quadratic equation 3x2 + (k − 1)x + 9 = 0 Putting x = 3 in above given eq. , we get 27 + 3(k − 1) + 9 = 0 ⇒ 27 + 3k - 3 + 9 = 0 ⇒ 3k = −33 ⇒ k = −11. Hence , the value of k is −11.

Question 13

One root of the equation 3x2 - 10x + 3 = 0 is 1 . Find the other root. 3

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Question 14

The expression x4 + 7x2 + 16 can be factored as :

Accepted Answer

The given expression x4 + 7x2 + 16 = ( x4 + 8x2 + 16 ) - x2 = ( x4 + 2 * 4 * x2 + 42 ) - x2 = ( x2 + 4 )2 - x2 = ( x2 + 4 + x ) ( x2 + 4 - x ) Hence , x4 + 7x2 + 16 = ( x2 + x + 4 ) ( x2 - x + 4 ). Option C is correct answer .

Question 15

The common root of the equations x2 - 7x + 10 = 0 and x2 - 10x + 16 = 0 is :

Accepted Answer

As per the given above question , we can say that The given quadratic equations are x2 - 7x + 10 = 0 ⇔ (x − 5)(x − 2) = 0 ⇔ x = 5, 2 And x2 - 10x + 16 = 0 ⇔ (x − 8)(x − 2) = 0 ⇔ x = 8, 2 ∴ Common root is 2.

Question 16

Divide 16 into 2 parts such the twice the square of the larger part exceeds the square of the smaller part by 164.

Accepted Answer

Let the smaller part be x. Then the larger part = 16 − x. According to question , we have 2( larger part )2 - ( smaller part )2 = 164 Now , 2( 16 - x )2 - x2 = 164 ⇒ 2( 256 + x2 - 32x ) - x2 = 164 ⇒ x2 - 64x + 348 = 0 ⇒ x2 - 6x - 58x + 348 = 0 ⇒ x(x - 6) - 58(x - 6) = 0 ⇒ (x - 6) (x - 58) x = 6 or x = 58 But m = 58 is not possible, since sum of the two parts is 16 ∴ smaller part ( x ) = 6, ∴ other part = 16 - x = 16 - 6 = 10. Hence , required answer will be option A .

Question 17

The solution of 2 - x = x - 2 would include : x

Accepted Answer

Given equation is 2x - x2 = x - 2 ⇔ x2 - x - 2 = 0 ⇔ ( x + 1 ) ( x - 2 ) = 0 ⇔ x = 2 or = - 1. Therefore , option B is correct answer .

Question 18

If log10 ( x2 - 6x + 45 ) = 2, then the values of x are

Accepted Answer

Given that : - log10 ( x2 - 6x + 45 ) = 2 ⇔ x2 - 6x + 45 = 102 = 100 ⇔ x2 - 6x - 55 = 0 ⇔ ( x - 11 ) ( x + 5 ) = 0 ⇔ x = 11 or x = - 5. Hence , the values of x are 11 and - 5.

Question 19

If the equations x2 + 2x - 3 = 0 and x2 + 3x - k = 0 have a common root, then the non - zero value of k is :

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Question 20

Find the quadratic equation whose roots are reciprocal of the roots of the equation 3x2 - 20x +17 = 0

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Quadratic Equation Questions