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Let $$f(x) = ax^2 + bx + c$$, where a, b and c are certain constants and $$a \neq 0$$ ?
It is known that $$f(5) = - 3f(2)$$. and that 3 is a root of $$f(x) = 0$$.
What is the other root of f(x) = 0?
f(3) = 9a + 3b + c = 0 f(5) = 25a + 5b + c
f(2) = 4a + 2b + c
f(5) = -3f(2) => 25a + 5b + c = -12a -6b -3c
=> 37a + 11b + 4c = 0 --> (1)
4(9a + 3b + c) = 36a + 12b + 4c = 0 --> (2)
From (1) and (2), a - b = 0 => a = b
=> c = -12a
The equation is, therefore, $$ax^2 + ax - 12a = 0 => x^2 + x - 12 = 0$$
=> -4 is a root of the equation.
Video Solution
Kamaldeep Singh
4 years, 11 months ago
Hi, How did the equation - 4(9a+3b+c) has come. kindly explain why there is a need to multiply 4 * f(3).
Devraj Patil
2 months, 1 week ago
Equation got from subtraction, has f(n+1) term on LHS but in next step question is solved by considering RHS only why f(n+1) term of LHS is not used?
