#17**+3 **

**I suspect this rude, careless dumb-dumb intended this equation:**

\(\LARGE x^2 - ax + b = 0\\ \)

Geometric solutions for this quadratic form are found by plotting A(0, 1) and B(a, b) and using the midpoint and distance formulas for finding (y) intercepts in circles. This seems consistent with the question. Examples of these are sometimes included in pre-calculus texts.

The first use of this form comes from Thomas Carlyle (1795–1881), known more for satirical social commentary than mathematics.

GA

GingerAleJan 20, 2019

#2**+3 **

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

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

MelodyJan 20, 2019

Jan 19, 2019