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If a,b are odd integers, the roots of the equation 2ax2+(2a+b)x+b=0, a≠0 are
- a)rational
- b)irrational
- c)non-real
- d)equal
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
If a,b are odd integers, the roots of the equation 2ax2+(2a+b)x+b=0, a...
We can first simplify the equation by factoring out a 2x:
2ax(2ax + 2b) - b = 0
Now we can use the zero product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. In this case, the two factors are:
2ax and 2ax + 2b - b/2a
Setting each factor equal to zero and solving for x, we get:
2ax = 0 -> x = 0 (since a is odd, this is a valid solution)
2ax + 2b - b/2a = 0 -> 4ax + 4b - b = 0 -> x = (-3b)/(4a)
So the roots of the equation are x = 0 and x = (-3b)/(4a).
2ax(2ax + 2b) - b = 0
Now we can use the zero product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. In this case, the two factors are:
2ax and 2ax + 2b - b/2a
Setting each factor equal to zero and solving for x, we get:
2ax = 0 -> x = 0 (since a is odd, this is a valid solution)
2ax + 2b - b/2a = 0 -> 4ax + 4b - b = 0 -> x = (-3b)/(4a)
So the roots of the equation are x = 0 and x = (-3b)/(4a).
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Community Answer
If a,b are odd integers, the roots of the equation 2ax2+(2a+b)x+b=0, a...
D = 4a^2 + b^2 + 4ab - 4 x 2a x b = 4a^2 + b^2 - 4ab = (2a - b)^2
so,
x = [-(2a + b) + or - (2a - b)] / 2a.
alpha = [-2a - b + 2a - b ]/2a. = -2b/2a = -b/a
beeta = [-2a - b - 2a + b]/2a = -4a/2a = -2
Here we can clearly see both roots are rational.Thus option (a) is the correct answer
so,
x = [-(2a + b) + or - (2a - b)] / 2a.
alpha = [-2a - b + 2a - b ]/2a. = -2b/2a = -b/a
beeta = [-2a - b - 2a + b]/2a = -4a/2a = -2
Here we can clearly see both roots are rational.Thus option (a) is the correct answer
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If a,b are odd integers, the roots of the equation 2ax2+(2a+b)x+b=0, a≠0 area)rationalb)irrationalc)non-reald)equalCorrect answer is option 'A'. Can you explain this answer?
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