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- Question
- ?. x 2 -24x + 144 = 0
?. y 2 ? 26y + 169 = 0
Options- A. 1
- B. 2
- C. 3
- D. 4
- Correct Answer
- 1
ExplanationAs per the given above question , we have
From equation ?. x2 - 24x + 144 = 0
? x2 ? 12x - 12x + 144 = 0
? x(x ? 12) ? 12(x ? 12) = 0
? (x ? 12)2 = 0
? x = 12
From equation ?. 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 . - 1. 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 :
Options- A. 4
- B. 6
- C. 2
- D. 8 Discuss
- 2. The roots of the equation 4x - 3 × 2x × 22 + 32 = 0 would include :
Options- A. 1, 2 and 3
- B. 1 and 2
- C. 1 and 3
- D. 2 and 3 Discuss
- 3. Consider the equation px2 + qx + r = 0, where p, q, r are real. The roots are equal in magnitude but opposite in sign when :
Options- A. q = 0, r = 0, p ? 0
- B. p = 0, qr ? 0
- C. r = 0, pr ? 0
- D. q = 0, pr ? 0 Discuss
- 4. If ?, ? are the roots of the equation x2 - 5x + 6 = 0, construct a quadratic equation whose roots are
1 , 1 . ? ?
Options- A. 6x2 + 5x - 1 = 0
- B. 6x2 - 5x - 1 = 0
- C. 6x2 - 5x + 1 = 0
- D. 6x2 + 5x + 1 = 0 Discuss
- 5. If x = ? ?(5 + 1)/?(5 - 1), then x 2 - x - 1 is equal to
Options- A. 0
- B. 1
- C. 2
- D. 5 Discuss
- 6. ?. 2x 2 + 3x ? 20 = 0
?. 2y 2 + 19y + 44 = 0
Options- A. 1
- B. 2
- C. 3
- D. 4 Discuss
- 7. ?. 10x 2 ? 7x + 1 = 0
?. 35y 2 -12y + 1 = 0
Options- A. 1
- B. 2
- C. 3
- D. 4 Discuss
- 8. ?. 6x 2 + 77x + 121 = 0
?. y 2 + 9y ? 22 = 0
Options- A. 1
- B. 2
- C. 3
- D. 4 Discuss
- 9. ?. x 2 ? 6x = 7
?. 2y 2 + 13y + 15 = 0
Options- A. 1
- B. 2
- C. 3
- D. 4 Discuss
- 10. In the following determine the set of value of P for which the given quadratic equation has real roots. Px2 + 4 x + 1 = 0
Options- A. P ? 4
- B. P > 4
- C. P ? 4
- D. P ? 4 Discuss
Quadratic Equation problems
Search Results
Correct Answer: 4
Explanation:
Correct Answer: 2 and 3
Explanation:
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.
Correct Answer: q = 0, pr ? 0
Explanation:
Correct Answer: 6x2 - 5x + 1 = 0
Explanation:
Correct Answer: 0
Explanation:
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
Correct Answer: 4
Explanation:
Correct Answer: 4
Explanation:
Correct Answer:
Explanation:
Correct Answer: 2
Explanation:
Correct Answer: P ? 4
Explanation:
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 .
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