Hi all,
Looking into the problem below and having a bit of trouble understanding why the answer is “4” solutions possible. The answer suggests that you use b^2 + 4ac to determine if there are 2 solutions (if the expression with numbers plugged in is positive), 1 solution (if the expression with numbers plugged in is zero) or no solutions (if the expression with numbers plugged in is negative).
Based on what I’ve been able to figure out arithmetically, the only possible solutions are -2 and -1. If you plug in 2 you get 10 and when you plug in 1 you get 4.
What other two numbers can you plug in??
How many real solutions are there for:
∣𝑥^2 + 3𝑥∣ = 2
Solutions:
0
1
2
4
infinite
Hi,
So, here, we will first consider two cases for opening the modulus.
Case 1: ∣𝑥^2 + 3𝑥∣ = 𝑥^2 + 3𝑥, given that 𝑥^2 + 3𝑥 >0 i.e. x(x+3) > 0
Here, 𝑥^2 + 3𝑥 = 2
D = 9 - 4(1)(-2) = 17 > 0. Therefore, 2 solutions here. Since D is not a perfect square, the solutions will be irrational.
Our solutions also satisfy x(x+3) > 0. Therefore, both solutions are valid.
(You can compare these values from the answer mentioned by Leaderboard in replies)
Case 2: ∣𝑥^2 + 3𝑥∣ = -(𝑥^2 + 3𝑥), given that 𝑥^2 + 3𝑥 <0 i.e. x(x+3) < 0
Here, -𝑥^2 - 3𝑥 = 2
So, 𝑥^2 + 3𝑥 + 2 = 0. By factorization, we get (x+2)(x+1)=0. There are two solutions here, too.
Our solutions also satisfy x(x+3) < 0, so these are both valid solutions.
So, the total solutions will be 4.
