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If L+M+N=0 and LMN are rationales the roots of the equation (M+N-L)x2+(N+L-M)x+(L+M-N)=0 are
- a)Real and irrational
- b)Real and rational
- c)Imaginary and equal
- d)Real and equal
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
If L+M+N=0 and LMN are rationales the roots of the equation (M+N-L)x2+...
Solution:
Given, L M N=0 and LMN are rationales
Let's assume roots of the equation to be x and y
We know that sum of roots = -b/a and product of roots = c/a
Therefore,
x+y = (N L-M)/(M N-L) + (L M-N)/(M N-L)
= (N L-M+L M-N)/(M N-L)
= -2MN/(M N-L)
xy = (N L-M)(L M-N)/(M N-L)
Given LMN is rational, therefore one of the roots is rational and the other is irrational.
Let's assume x is rational, then y will be irrational.
Sum of rational and irrational is always irrational. Therefore, x+y is irrational.
Product of rational and irrational is always irrational. Therefore, xy is irrational.
Therefore, option 'a' and 'c' are incorrect.
So, the correct option is 'b', i.e. Real and rational.
Given, L M N=0 and LMN are rationales
Let's assume roots of the equation to be x and y
We know that sum of roots = -b/a and product of roots = c/a
Therefore,
x+y = (N L-M)/(M N-L) + (L M-N)/(M N-L)
= (N L-M+L M-N)/(M N-L)
= -2MN/(M N-L)
xy = (N L-M)(L M-N)/(M N-L)
Given LMN is rational, therefore one of the roots is rational and the other is irrational.
Let's assume x is rational, then y will be irrational.
Sum of rational and irrational is always irrational. Therefore, x+y is irrational.
Product of rational and irrational is always irrational. Therefore, xy is irrational.
Therefore, option 'a' and 'c' are incorrect.
So, the correct option is 'b', i.e. Real and rational.
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Community Answer
If L+M+N=0 and LMN are rationales the roots of the equation (M+N-L)x2+...
As M,L and N are rationals, any addition or subtraction among them will always render a rational number.Hence option B is right
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If L+M+N=0 and LMN are rationales the roots of the equation (M+N-L)x2+(N+L-M)x+(L+M-N)=0 area)Real and irrationalb)Real and rationalc)Imaginary and equald)Real and equalCorrect answer is option 'B'. Can you explain this answer?
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