Discussion
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- Question
- The equation x2 ? px + q = 0, p, q ? R has on real root if :
Options- A. p2 ? 4q
- B. p2 < 4q
- C. p2 > 4q
- D. None of these
- Correct Answer
- p2 < 4q
ExplanationAccording 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 . - 1. ?. 5x + 2y = 31
?. 3x + 7y = 36
Options- A. 1
- B. 2
- C. 3
- D. 4 Discuss
- 2. ?. x 2 - 7x + 12 = 0
?. y 2 - 12y + 32 = 0
Options- A. 1
- B. 2
- C. 3
- D. 4 Discuss
- 3. ?. 2x 2 + 11x + 14 = 0
?. 4y 2 + 12y + 9 = 0
Options- A. 1
- B. 2
- C. 3
- D. 4 Discuss
- 4. ?. x 2 = 729
?. y = ? 729
Options- A. 1
- B. 2
- C. 3
- D. 4 Discuss
- 5. ?. x 2 ? 4 = 0
?. y 2 + 6y + 9 = 0
Options- A. 1
- B. 2
- C. 3
- D. 4 Discuss
- 6. x = 3a solution of the equation 3x2 + (k ? 1)x + 9 = 0 if k has value
Options- A. 13
- B. ?13
- C. 11
- D. ?11 Discuss
- 7.
One root of the equation 3x2 - 10x + 3 = 0 is 1 . Find the other root. 3
Options- A. 3
- B. 1 3
- C. - 3
- D. None of these Discuss
- 8. The expression x4 + 7x2 + 16 can be factored as :
Options- A. (2 + x + 1) (2 + x + 16)
- B. (2 + x + 1) (2 ? x + 16)
- C. (2 + x + 4) (2 ? x + 4)
- D. (2 + x ? 4) (2 ? x ? 4) Discuss
- 9. The common root of the equations x2 - 7x + 10 = 0 and x2 - 10x + 16 = 0 is :
Options- A. - 2
- B. 3
- C. 5
- D. 2 Discuss
- 10. Divide 16 into 2 parts such the twice the square of the larger part exceeds the square of the smaller part by 164.
Options- A. 10, 6
- B. 8, 8
- C. 12, 4
- D. None of these Discuss
Quadratic Equation problems
Search Results
Correct Answer: 1
Explanation:
Given that :- Given that :- ?. 5x + 2y = 31 .....( 1 )
?. 3x + 7y = 36 .....( 2 )
Solving these two linear equations, we get x = 5, y = 3 .
? x > y is correct answer .
Correct Answer: 4
Explanation:
As per the given above equations , we have
From equation ?. 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
?. 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 .
Correct Answer: 3
Explanation:
Correct Answer: 4
Explanation:
According to question ,we have
From equation ?. x2 = 729
? x = ± ?729 = ± 27
From equation ?. y = ?729
? y = 27
Hence , x ? y is required answer .
Correct Answer: 1
Explanation:
According to question ,we have
From equation ?. x2 ? 4 = 0
? x = ± 2,
From equation ?. 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 .
Correct Answer: ?11
Explanation:
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.
Correct Answer: 3
Explanation:
Correct Answer: (2 + x + 4) (2 ? x + 4)
Explanation:
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 .
Correct Answer: 2
Explanation:
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.
Correct Answer: 10, 6
Explanation:
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 .
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