Math Module 2 (Hard) Question 15

In the given equation, $c$ is a constant. For which of the following values of $c$ will the equation have more than two real solutions?
$$x^2(x+1)+4x(x+1)+c(x+1)=0$$
Answer:
A

Explanation

The correct answer is Choice A.

Factor the given equation to determine the number of $x$-intercepts, to find the value of $c$ such that the given equation has more than two solutions.

Therefore, $c=2$ is the only value from the given choices for the equation to have more than two solutions.

Choice B is incorrect and may result from finding a value that gives exactly two real solutions.

Choices C and D are incorrect and may result from algebraic mistakes.

✨ Expert's Tip ✨

When factored into a perfect squared, the solution derived from the factor only counts as one real solution.

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