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# help

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

Jan 19, 2019

#1
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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!

Jan 20, 2019
#2
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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

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

Jan 20, 2019
edited by Melody  Jan 20, 2019