$x^2-2(a+1)x+3a+2=0$
Where a is any real number. Find all values of a so that the equation has two distinct real roots.
I tried solving for a by using the inequality $b^2-4ac\gt0$... but when i substitute the value into the equation and input it into a graphing calculator, there are no distinct roots...
On
Hint: If $Ax^2+Bx+C=0$ then $$x=\frac{-B\pm\sqrt{B^2-4AC}}{2A}$$
note that $A=1$, $B=-2(a+1)$ and $C=3a+2$. Now you solve the inequality:
$$B^2-4AC>0$$
On
The only pairs of $(a,x)$ that seem to work for $\quad x^2-2(a+1)x+3a+2=0\quad$ are
$$\quad (-1,\pm1)\quad (2,2)\quad (2,4)\quad \bigg(\frac{1 - \sqrt{5}}{2},\frac{3 - \sqrt{5}}{2}\bigg)\quad \bigg(\frac{1 + \sqrt{5}}{2},\frac{3 + \sqrt{5}}{2}\bigg)$$ You can confirm this integer values in a spreadsheet if you test for $\quad-10\le a \le 10$ $$x = \pm\sqrt{a^2 - a - 1} + a + 1$$
Beginning with $x^2-2(a+1)x+3a+2=0$, we wish to have the canonical-form $b^2-4ac\ge 0$. This looks like:
$$4(a+1)^2-4(3a+2)\ge 0\\ a^2+2a+1-3a-2\ge 0\\ a^2-a-1\ge 0\\ a=\frac {1\pm \sqrt{1+4}}2=\frac 12\pm\frac {\sqrt5} 2$$
So either $a\gt \frac 12+\frac {\sqrt 5}2$ or $a\lt \frac 12-\frac {\sqrt 5}2$. Did you plug in those values for $a$? At $a=\frac 12\pm\frac{\sqrt 5}2$, you will get a single real solution of multiplicity $2$.
Now consider a specific number, say $a=3\gt \frac 12+\frac{\sqrt 5}2$. Then we have $x^2-8x+11=0,$ with real solutions at $x=4\pm\sqrt{5}.$ For another example, consider $a=-1\lt\frac 12-\frac{\sqrt 5}2$ where we get $x^2-1=0$ with solutions $x=-1,1$.