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- Question
- If ? and ? are the roots of the equation x 2 - 11x + 24 = 0, find the equation having the roots ? + 2 and ? + 2
Options- A. x2 + 15x + 24 = 0
- B. x2 - 15x + 24 = 0
- C. x2 + 15x - 50 = 0
- D. x2 + 15x - 60 = 0
- Correct Answer
-
ExplanationGiven equation is
x2 - 11x + 24 = 0
Then, required equation is
(x - 2)2 - 11(x - 2) + 24 = 0
? x2 - 4x + 4 - 11x + 22 + 24 = 0
? x2 - 15x + 50 = 0 - 1. If ? and ? be the roots of the equation ax 2 + bx + c = 0, find the value of ? 2/? + ? 2/? .
Options- A. ab -b2c/2b2c
- B. 3bc - a3/b2c
- C. 3ac - b2/a3c
- D. 3abc - b3/a2c Discuss
- 2. If a and b are the roots of the equation x 2 - 6x + 6 = 0, find the value of 2(a 2 + b 2).
Options- A. 40
- B. 42
- C. 48
- D. 46 Discuss
- 3. If ? and ? are the roots of the equation x2 - 3?x + ?2 = 0,
find ? if ?2 + ?2 = 7 4
Options- A. ± 1 2
- B. ± ?7 2
- C. ± ?3 2
- D. None of these Discuss
- 4. Determine k such that the quadratic equation x2 - 2( 1 + 3k ) x + 7 ( 3 + 2k ) = 0 has equal roots.
Options- A. 2, - 10 9
- B. 2, 10 9
- C. - 2, 10 9
- D. - 2, -10 9 Discuss
- 5. If ?, ? are the roots of the equation x2 + kx + 12 = 0 such that ? ? ? = 1, the value of k is
Options- A. 0
- B. ± 5
- C. ± 1
- D. ± 7 Discuss
- 6. If x 2 = 6 + ? 6 + ?6 + ?6 + .... ? , then what is one of the value of x equal to?
Options- A. 6
- B. 5
- C. 4
- D. 3 Discuss
- 7. If a = p / (p + q) and b = q / (p - q), then and ab/a + b is equal to
Options- A. pq / (p2 + q2)
- B. (p2 + q2) / pq
- C. p / (p + q)
- D. [p / (p + q)]2 Discuss
- 8. The sum of the roots of the equation x 2 + px + q = 0 is equal to the sum of their squares, then
Options- A. p2 - q2 = 0
- B. p2 + q2 = 2q
- C. p2 + p = 2q
- D. q2 + q2 = 2p Discuss
- 9. If ? and ? are the roots of the equation 8x 2 - 3x + 27 = 0, find the value of (? 2/?) 1/3 + (? 2/?) 1/3
Options- A. 1/3
- B. 1/4
- C. 7/2
- D. 4 Discuss
- 10. 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 equation was
Options- A. x2 + 10x + 9 = 0
- B. x2 - 10x + 16 = 0
- C. x2 - 10x + 9 = 0
- D. None of the above Discuss
Quadratic Equation problems
Search Results
Correct Answer: 3abc - b3/a2c
Explanation:
Given, ? and ? are the roots of the equation ax2 + bx + c = 0
? Sum of two roots = -b/a
? + ? = -b/a
and product of two roots = c/a
? ? = c/a
? ?2/? + ?2/? = ( ?3 + ?3) / ? ?
= [(? + ?) 3 - 3 ? ? (? + ?)] / ??
[? (a + b)3 = a3 + b3 + 3ab(a + b) ]
= (-b/a)3 - [3c/a(-b/a)/c/a = -b3/a3 + 3bc/a2]/(c/a)
= 3abc - b3/a2c
Correct Answer: 48
Explanation:
a and b are the roots of the equation
x2 - 6x + 6 = 0
? a + b = -B/A = 6 and ab = C/A = 6
We know that,
a2 + b2 = (a + b)2 - 2ab
= (6)2 - 2 x 6 = 36 - 12 = 24
? 2(a2 + b2) = 2 x 24 = 48
Correct Answer: ± 1 2
Explanation:
Correct Answer: 2, - 10 9
Explanation:
Correct Answer: ± 7
Explanation:
As per the given above question , we know that
Let, ?, ? be the roots of the equation x2 + kx + 12 = 0
? ? + ? = ?k and ?? = 12
Now ( ? - ? )2 = ( ? + ? )2 - 4??
( 1 )2 = k2 - 4(12)
? k2 = 49 ? k = ± 7.
Correct Answer: 3
Explanation:
Here, x2 = 6 + ?6 + ?6 + ?6 + .... ? ,
So, x2 = 6 + ?x2
? x2 = 6 + x
? x2 - x - 6 = 0
? x2 + 2x - 3x - 6 = 0
? x(x + 2) - 3(x + 2) = 0
? (x - 3) (x + 2) = 0
? x = 3
Correct Answer: pq / (p2 + q2)
Explanation:
ab/(a + b) = [p/(p + q)] x [q/(p - q)] / [p/(p + q) + q/(p - q)]
= pq/p2 - pq + pq + q2
= pq/(p2 + q2)
Correct Answer: p2 + p = 2q
Explanation:
Let ? and ? be the roots of the equation
x2 + px + q = 0
Then, ? + ? = -p, ? ? = q
According to the question,
? + ? = ?2 + ?2
? ? + ? = (? + ?)2 - 2? ?
? -p = p2 - 2q ? p2 + p = 2q
Correct Answer: 1/4
Explanation:
Since, ? and ? are the roots of the equation
8x2 - 3x + 27 = 0
? ? + ? = 3/8 and ? ? = 27/8
? (?2/?)1/3 + (?2/? )1/8
= (?3)1/3 + (?3)1/3/(??)3
= ? + ? / (??)1/3 = 3/8/(27/8)1/3
= 3/8 x 2/3 = 1/4
Correct Answer: x2 - 10x + 9 = 0
Explanation:
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.
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