Determine all real numbers a such that the inequality \( |x^2 + 2ax + 3a|\le2\) has exactly one solution in x.
+532 a can be anything, because x can be a number, but there is always another solution because there are no restrictions on x.
HOPE THIS HELPED!
+118608 Determine all real numbers a such that the inequality \(|x^2 + 2ax + 3a|\le2\) has exactly one solution in x.
From the graph I have draw in Desmos I can see that this has exaclty 2 solutions. a=2 and a=1
https://www.desmos.com/calculator/fewruikgqj
\(|x^2 + 2ax + 3a|\le2\\ |x^2 + 2ax + 3a|-2\le0\)
Take a look at the graph and see if you can work out what it is showing you.
Now I will talk it through algebraically:
Now \(y=x^2+2ax+3a\) is a concave up parabola. When it is put in absolute signs the bit in the middle, that was under the x axis is reflected in the x axis. So effectively this means that the bit between the roots becomes a concave down parabola.
When this is dropped by 2 units there will be either 2 or 4 roots. So \(x^2+2ax+3a \) must be positive (because there can only be one root.)
\(|x^2 + 2ax + 3a|=2\\ \qquad x^2+2ax+3a>0 \qquad so\\ x^2+2ax+3a=2\\ x^2+2ax+3a-2=0\\ x^2+2ax+(3a-2)=0 \\ \text{I only want one solution so the discriminant must be 0}\\ (2a)^2-4*1*(3a-2)=0 \\ 4a^2-12a+8=0 \\ a=\frac{12\pm\sqrt{144-128}}{8}\\ a=\frac{12\pm\sqrt{16}}{8}\\ a=\frac{12\pm4}{8}\\ a=\frac{16}{8}\:\;or\;\frac{8}{8}\\ a=2\:\;or\;1\)
Hence
\(When\;\;a=1\\ |x^2 + 2x + 3|\le2\\ \text{is the same as } \quad x^2 + 2x + 3 \le2\\ \text {And it has only one solution }x=-1\)
and
\(When\;\;a=2\\ |x^2 + 4x + 6|\le2\\ \text{is the same as } \quad x^2 + 4x + 6 \le2\\ \text {And it has only one solution }x=-2\)
