Question

f(x) = x^2 + 6x - 16
two real rational solutions
g(x) = x^2 +6x + 1
two real irrationals solutions

i found that f(x) is two real rational solutions and that g(x) is two real irrationals solutions on another open study question but dont understand how to get that

Answer 1

@kirbykirby

Answer 2

You can apply the quadratic formula:
If you have a polynomial of the form \(ax^2+bx+c\), then \[ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

For \(f(x) = x^2 + 6x - 16 \), \(a=1, b=6, c=-16\)
\[ x=\frac{-6\pm\sqrt{6^2-4(1)(-16)}}{2(1)}\\
x=\frac{-6\pm\sqrt{100}}{2}\\
x=\frac{-6\pm 10}{2}\\
x=\frac{-6-10}{2}, or ~ x=\frac{-6+10}{2}\]

For \(g(x) = x^2 +6x + 1\), \(a=1, b=6, c=1\)
\[ x=\frac{-6\pm\sqrt{6^2-4(1)(1)}}{2(1)}\\
x=\frac{-6\pm\sqrt{32}}{2}\\

x=\frac{-6-\sqrt{32}}{2}, or ~ x=\frac{-6+\sqrt{32}}{2}\]
But \(\sqrt{32}=4\sqrt{2}\) and is irrational because \(\sqrt{2}\) is irrational. So thus, the roots above will also be irrational.

Answer 3

@kirbykirby how would i convert f(x) into general, vertex form

Answer 4

This would involve the process of completing the square.
For \( x^2 + 6x - 16\), first take the \(b\) coefficient (\(6\)), and add and subtract \(\left(\dfrac{b}{2}\right)^2\), so \(\left(\dfrac{6}{2}\right)^2=9\)
Thus:
\[ x^2+6x\color{red}{+9-9}-16\]
Now the first 3 terms will give you a binomial of the form \((a+b)^2\). Since this is equal to \(a^2+2ab+b^2\), then it follows from the 1st three terms above that \(a=1\) and \(b=3\)
\[ (x+3)^2-25\], the -25 comes from combining -9 and -16
So it is in vertex form \(a(x-h)^2+k\) where \(a=1, h=-3, k=-25\)