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- Question
- If one of the roots of the equation x 2 - bx + c = 0 is the square of the other, then which of the following option is correct?
Options- A. b3 = 3bc + c2 + c
- B. c3 = 3bc + b2 + b
- C. 3bc = c3 + b2 + b
- D. 3bc = c3 + b3 + b2
- Correct Answer
- b3 = 3bc + c2 + c
ExplanationGiven that, one root of the equation
x2 - bx + c = 0 is square of other root of this equation i.e., roots (?, ?2).
? Sum of roots = ? + ?2 = -(-b)/1
? ? (? + 1) = b .......(i)
and product of roots= ?. ?2 = c/1
? ?3 = c ? = c1/3 ....(ii)
From Eqs. (i) and (ii).
c1/3 (c1/3 + 1) = b ....(iii)
On cubing both sides, we get
c(c1/3 + 1)3 = b3
? c {c + 1 + 3c1/3 (c1/3 + 1)} = b3
? c {c + 1 = 3b} = b3 [from Eq. (iii)]
? b3 = 3bc + c2 + c - 1. 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
- 2. 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
- 3. 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
- 4. 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
- 5. 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
- 6. Two students A and B solve an equation of the from x 2 + px + q = 0. A starts with a wrong value of p and obtains the roots as 2 and 6. B starts with a wrong value of q and gets the roots as 2 and -9. What are the correct roots of the equation?
Options- A. 3 and - 4
- B. - 3 and - 4
- C. - 3 and 4
- D. 3 and 4 Discuss
- 7. If x = ? ?(5 + 1)/?(5 - 1), then x 2 - x - 1 is equal to
Options- A. 0
- B. 1
- C. 2
- D. 5 Discuss
- 8. 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
- 9. 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
- 10. 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
Quadratic Equation problems
Search Results
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.
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: 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: 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: 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: - 3 and - 4
Explanation:
Let ? and ? be the roots of the quadratic equation x2 + px + q = 0
Given that, A starts with a wrong value of p and obtains the roots as 2 and 6. But this time q is correct. i.e products of roots
= q = ? . ? = 6 x 2 = 12 ...(i)
and B starts with a wrong value of q and gets the roots as 2 and - 9. But this time p is correct i.e., sum of roots
= p = ? + ? = - 9 + 2 = - 7 ..(ii)
(? - ?)2 = (? + ?)2 - 4? ?
= (-7)2 - 4.12 = 49 - 48 = 1
[from Eqs. (i) and (ii)]
? ? - ? = 1 ..(iii)
Now, from Eqs. (ii) and (iii) , we get
? = - 3 and ? = - 4
which are correct roots .
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: 6x2 - 5x + 1 = 0
Explanation:
Correct Answer: q = 0, pr ? 0
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.
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