Quadratic Equation problems
- 1. If p and q are the roots of the equation x 2 + px + q = 0, then
Options- A. p = 1 and q = -2
- B. p = 0 and q = 1
- C. p = -2 and q = 0
- D. p = -2 and q = 1 Discuss
Correct Answer: p = 1 and q = -2
Explanation:
Since, p and q are the roots of the equation x2 + px + q = 0
Then, p + q = - p
and pq = q
Now, pq = q
? p = 1
Putting the value of p in p + q = - p, we get
1 + q = - 1
? q = - 2
- 2. Number of solution of the equation ? x2 - x + 1 + 1/? x2 - x + 1 = 2 - x 2 is
Options- A. 0
- B. 1
- C. 2
- D. 4 Also asked in: Bank Exams
Correct Answer: 1
Explanation:
We know that, AM ? GM
? ?a + 1/?a ? 2
Here, ?x2 - x + 1 + 1/?x2 - x + 1 ? 2
? 2 - x2 ? 2
? x2 ? 0
? x = 0
Hence, the given equation has only one solution.
- 3. Two number whose sum is 8 and difference is 4, are the roots of the equation
Options- A. x2 + 8x + 12 = 0
- B. x2 + 8x - 12 = 0+
- C. x2 - 8x - 12 = 0
- D. x2 - 8x + 12 = 0 Discuss
Correct Answer: x2 - 8x + 12 = 0
Explanation:
Let the roots be ? and ? .
Then, ? + ? = 8 ......(i)
? - ? = 4 ..(ii)
On solving Eqs. (i) and (ii), we get
? = 6, ? = 2
? Required equation is
x2 - (? + ?) x + (? ?) = 0
? x2 - (6 + 2)x + 6 x 2 = 0
? x2 - 8x + 12 = 0
- 4. The sum of a number and its reciprocal is 10/3, then the numbers are
Options- A. 3, 1/3
- B. 3, -1/3
- C. -3, 1/3
- D. -3, -1/3 Discuss
Correct Answer: 3, 1/3
Explanation:
Let the number be x and 1/x,
Then, x + 1/x = 10/3
? (x2 + 1)/x = 10/3
? 3x2 - 10x + 3 = 0
? 3x2 - 9x - x + 3 = 0
? 3x(x - 3) -1 (x - 3) = 0
? (3x - 1) (x - 3) = 0
? x = 1/3, x = 3
- 5. The values of p for which the difference between the roots of the equation x 2 + px + 8 = 0 is 2 , are
Options- A. ± 2
- B. ± 4
- C. ± 6
- D. ± 8 Discuss
Correct Answer: ± 6
Explanation:
? + ? = - p, ? ? = 8
Given, ? - ? = 2
On squaring both sides, we get
(? - ?)2 = 4
? (? + ?)2 - 4?? = 4
? p2 - 32 = 4
? p2 = 36 ? p = ± 6
- 6. x 2 + x - 20 = 0
y 2 - y - 30 = 0
Options- A. If x > y
- B. If x ? y
- C. If x < y
- D. If x ? y Discuss
Correct Answer: If x ? y
Explanation:
x2 + x - 20 = 0
[by factorisation method]
? x2 + 5x - 4x - 20 = 0
? x(x + 5) - 4(x + 5) = 0
(x + 5) (x - 4) = 0
? x = -5 or 4
and y2 - y - 30 = 0
? y2 - 6y + 5y - 30 = 0
? y(y - 6) + 5(y - 6) = 0
? (y - 6) (y + 5) = 0
? y = 6 or - 5
Hence, y ? x or x ? y
- 7. 225x 2 - 4 = 0; ? 225y + 2 = 0
Options- A. If x > y
- B. If x ? y
- C. If x < y
- D. If x ? y Discuss
Correct Answer:
Explanation:
225x2 - 4 = 0;
? 225x2 = 4 ? x2 = 4/225
? x = ?4/225 = ± 2/15, i.e., 2/15 and -2/15
and ?225y + 2 = 0 or ?225y = -2
On squaring both sides, we get
(?225y)2 = (-2)2
? 225y = 4
? y = 4/225
So, relation cannot be established because 4/225 lies between 2/15 and -2/15.
- 8. x - ? 121 = 0
y 2 - 121 = 0
Options- A. If x > y
- B. If x ? y
- C. If x < y
- D. If x ? y Discuss
Correct Answer:
Explanation:
x - ?121 = 0
? x = ?121 ? x = ± 11
and y2 - 121 = 0 ? y2 = 121
? y = ?121 = ± 11
? x = y
- 9. x 2 - 16 = 0 y 2 - 9y + 20 = 0
Options- A. if x > y
- B. if x ? y
- C. if x < y
- D. if x ? y Discuss
Correct Answer: if x ? y
Explanation:
x2 - 16 = 0
? x2 = 16 ? x = ?16 = ± 4
and y2 - 9y + 20 = 0
y2 - 4y - 5y + 20 = 0
? y(y - 4) - 5(y - 4) = 0
? y = 5, 4
? y ? x or x ? y
- 10. x 2 - 32 = 112; y - ? 169 = 0
Options- A. If x > y
- B. If x ? y
- C. If x < y
- D. If x ? y Discuss
Correct Answer:
Explanation:
x2 - 32 = 112
? x2 = 112 + 32 ? x2 = 144
? x = ± 12
and y - ?169 = 0
? y = ?169
? y = ± 13
? Relation cannot be established.
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