Let a,b and c be real numbers such that 4a+2b+c =0 and ab>0. Then the eqn ax²+bx²+cx²=0 has-
To find the solution to the equation ax² + bx² + cx² = 0 with the given conditions, we can use the fact that the coefficients of x² are 4a, 2b, and c.
Given that 4a + 2b + c = 0, we can rewrite it as:
4a = -2b - c
Now, substitute this expression into the equation ax² + bx² + cx² = 0:
(-2b - c)x² + bx² + cx² = 0
Combine like terms:
(-2b + b + c)x² = 0
Simplify:
(-b + c)x² = 0
Now, we know that ab > 0. Since a and b have the same sign, we have two possibilities:
1. a > 0 and b > 0
2. a < 0 and b < 0
Case 1: a > 0 and b > 0
In this case, (-b + c) > 0, because -b and c have opposite signs. Therefore, for (-b + c)x² = 0 to be true, x² must be equal to zero. This means x = 0.
Case 2: a < 0 and b < 0
In this case, (-b + c) < 0, because -b and c have the same sign. Therefore, for (-b + c)x² = 0 to be true, x² must be equal to zero. This means x = 0.
In both cases, the only solution to the equation ax² + bx² + cx² = 0 is x = 0.