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$x^2-2(a+1)x+3a+2=0$

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...

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  • $\begingroup$ What inequality do you get when creating $b^2-4ac\ge 0$? $\endgroup$ – abiessu Sep 10 '14 at 23:37
  • $\begingroup$ @abiessu so this is what I did: [-2(a+1)]^2 - 4(1)(3a+2) >= 0 4a^2+8a+4-12a-8 >= 0 4a^2-4a-4 >= 0 and then i used the quadratic equation to find a and the answer i got it [1+/- (5)^0.5]/2 $\endgroup$ – mathemator Sep 10 '14 at 23:40
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    $\begingroup$ So far right. At the two roots $\sigma\lt \tau$ of $a^2-a-1=0$ we will have two coinciding solutions. For $a\lt \sigma$ or $a\gt \tau$ we will have two distinct real roots. $\endgroup$ – André Nicolas Sep 10 '14 at 23:45
  • $\begingroup$ I put an example solution at the bottom of my answer. Does this help clarify? $\endgroup$ – abiessu Sep 10 '14 at 23:49
  • $\begingroup$ yes it does! thank you $\endgroup$ – mathemator Sep 10 '14 at 23:52
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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:

$$B^2-4AC>0$$

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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 $x=-1,1$.

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