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If the roots of the equation x3 + Px2 + Qx – 19 = 0 are each one more than the roots of the equation x3 – Ax2 + Bx – C = 0, where A, B, C, P & Q are constants, then the value of A + B + C is equal to
- a)18
- b)19
- c)20
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
If the roots of the equation x3+ Px2+ Qx –19 = 0 are each one mo...
Let the roots of x^3−Ax^2 + Bx − C=0 be α,β,γ
⇒ Now, α + β + γ = −Coeffx^2 / Coeffx3
= −(−A) / 1 =A ---- ( 1 )
⇒ αβ + βγ + αγ = Coeff.ofx / Coeff.x3 = B ------- ( 2 )
⇒ αβγ = Coeff.1 / Coeff.x^3 =C -------- ( 3 )
⇒ According to the question, roots of the equation x3 + px2 + q − 19 = 0 are (α+1),(β+1),(γ+1)
⇒ As in the above example −p=α+1+β+1+γ+1=α+β+γ+3
19 = (α+1)(β+1)(γ+1) [ Product of roots ]
19 = αβγ + αβ + βγ + αγ + α + β + γ + !
19 = A+B+C+1 [ From ( 1 ), ( 2 ) and ( 3 ) ]
∴ A+B+C=18
Most Upvoted Answer
If the roots of the equation x3+ Px2+ Qx –19 = 0 are each one mo...
To find the roots of the equation, we can use the fact that the sum of the roots of a cubic equation is equal to the negation of the coefficient of the x^2 term divided by the coefficient of the x^3 term.
Therefore, the sum of the roots is -P/1 = -P.
Since the equation is symmetric, the product of the roots is equal to the constant term divided by the coefficient of the x^3 term.
Therefore, the product of the roots is Q/1 = Q.
We are given that the sum of the roots is 0 and the product of the roots is -Q.
Therefore, we can set up the following equations:
-P = 0
Q = -Q
From the first equation, we can solve for P:
P = 0
From the second equation, we can solve for Q:
Q = 0
Therefore, the roots of the equation are P = 0 and Q = 0.
Therefore, the sum of the roots is -P/1 = -P.
Since the equation is symmetric, the product of the roots is equal to the constant term divided by the coefficient of the x^3 term.
Therefore, the product of the roots is Q/1 = Q.
We are given that the sum of the roots is 0 and the product of the roots is -Q.
Therefore, we can set up the following equations:
-P = 0
Q = -Q
From the first equation, we can solve for P:
P = 0
From the second equation, we can solve for Q:
Q = 0
Therefore, the roots of the equation are P = 0 and Q = 0.
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If the roots of the equation x3+ Px2+ Qx –19 = 0 are each one more than the roots of the equation x3–Ax2+ Bx –C = 0, where A, B, C, P & Q are constants, then the value of A + B + C is equal toa)18b)19c)20d)None of theseCorrect answer is option 'A'. Can you explain this answer?
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If the roots of the equation x3+ Px2+ Qx –19 = 0 are each one more than the roots of the equation x3–Ax2+ Bx –C = 0, where A, B, C, P & Q are constants, then the value of A + B + C is equal toa)18b)19c)20d)None of theseCorrect answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If the roots of the equation x3+ Px2+ Qx –19 = 0 are each one more than the roots of the equation x3–Ax2+ Bx –C = 0, where A, B, C, P & Q are constants, then the value of A + B + C is equal toa)18b)19c)20d)None of theseCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If the roots of the equation x3+ Px2+ Qx –19 = 0 are each one more than the roots of the equation x3–Ax2+ Bx –C = 0, where A, B, C, P & Q are constants, then the value of A + B + C is equal toa)18b)19c)20d)None of theseCorrect answer is option 'A'. Can you explain this answer?.
If the roots of the equation x3+ Px2+ Qx –19 = 0 are each one more than the roots of the equation x3–Ax2+ Bx –C = 0, where A, B, C, P & Q are constants, then the value of A + B + C is equal toa)18b)19c)20d)None of theseCorrect answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If the roots of the equation x3+ Px2+ Qx –19 = 0 are each one more than the roots of the equation x3–Ax2+ Bx –C = 0, where A, B, C, P & Q are constants, then the value of A + B + C is equal toa)18b)19c)20d)None of theseCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If the roots of the equation x3+ Px2+ Qx –19 = 0 are each one more than the roots of the equation x3–Ax2+ Bx –C = 0, where A, B, C, P & Q are constants, then the value of A + B + C is equal toa)18b)19c)20d)None of theseCorrect answer is option 'A'. Can you explain this answer?.
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