We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

Determine all real numbers a such that the inequality \( |x^2 + 2ax + 3a|\le2\) has exactly one solution in x.

 Jan 19, 2019

a can be anything, because x can be a number, but there is always another solution because there are no restrictions on x.



 Jan 20, 2019

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



\(|x^2 + 2ax + 3a|\le2\\ |x^2 + 2ax + 3a|-2\le0\)


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.)


\(|x^2 + 2ax + 3a|=2\\ \qquad x^2+2ax+3a>0 \qquad so\\ x^2+2ax+3a=2\\ x^2+2ax+3a-2=0\\ x^2+2ax+(3a-2)=0 \\ \text{I only want one solution so the discriminant must be 0}\\ (2a)^2-4*1*(3a-2)=0 \\ 4a^2-12a+8=0 \\ a=\frac{12\pm\sqrt{144-128}}{8}\\ a=\frac{12\pm\sqrt{16}}{8}\\ a=\frac{12\pm4}{8}\\ a=\frac{16}{8}\:\;or\;\frac{8}{8}\\ a=2\:\;or\;1\)




\(When\;\;a=1\\ |x^2 + 2x + 3|\le2\\ \text{is the same as } \quad x^2 + 2x + 3 \le2\\ \text {And it has only one solution }x=-1\)



\(When\;\;a=2\\ |x^2 + 4x + 6|\le2\\ \text{is the same as } \quad x^2 + 4x + 6 \le2\\ \text {And it has only one solution }x=-2\)


 Jan 20, 2019
edited by Melody  Jan 20, 2019

34 Online Users