I'm trying to solve the following equation $2t^2 + t - 3 = 0$
I start by dividing by 2, $t^2 + \frac {t}{2} - \frac {3}{2} = 0$
Then I solve for t $t = - \frac{ \frac {1}{2} }{2} \binom{+}{-} \sqrt{(\frac {1}{2})^2 + \frac {3}{2}}$
$t = - \frac{1}{4} \binom{+}{-} \sqrt{(\frac {1}{4}) + \frac {6}{4}}$
I calculate $t = - \frac{1}{4} \binom{+}{-} \frac {\sqrt7}{4}$
$t_1 = - \frac{1}{4} + \frac {\sqrt7}{4}$
$t_2 = - \frac{1}{4} - \frac {\sqrt7}{4}$
But according to wolframalpha it's suppose to be
$t_1 = 1$
$t_2 = - \frac {3}{2}$
Can't figure out where did I go wrong in my calculation?
$\endgroup$ 23 Answers
$\begingroup$Well actually you have applied the quadratic formula wrong. The roots of the equation $$ax^2+bx+c=0$$ is given by
$$\alpha, \beta ={-b \pm \sqrt {b^2-4ac} \over 2a}$$
So for the equation $$t^2+ \frac12t - \frac32=0$$ the roots become $$\alpha, \beta={- \frac12 \pm \sqrt { (\frac12)^2+4(\frac32)} \over 2}$$
which gives the roots as $1, -1.5$
But in cases like these, I'd suggest an alternative method. the equation can be re-written as $$2t^2-2t+3t-3=0$$
which can be written as $$2t(t-1)+3(t-1)=0$$
Now on taking $t-1$ as the common factor, we can write $$(t-1)(2t+3)=0$$
which gives the desired roots.
$\endgroup$ $\begingroup$we get $t_{1,2}=-\frac{1}{4}\pm\sqrt{\frac{1}{16}+\frac{24}{16}}$ and we get $t_1=-\frac{1}{4}+\frac{5}{4}=\frac{4}{4}=1$ and $t_2=-\frac{1}{4}-\frac{5}{4}=-3/2$
$\endgroup$ $\begingroup$In general, to solve $$ ax^2+bx+c=0$$ We can use the quadratic formula, which is $$ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} $$ In your case, we have $$ 2t^2+t-3=0$$ Which implies that $$ a=2, b=1, c=-3 $$ Also I'd recommend against your first step of dividing by $2$. This step introduces fractions and the quadratic formula is much simpler when just using integers. So now $$ t=\frac{-1\pm \sqrt{1^2-4(2)(-3)}}{2(2)} $$ $$ t=\frac{-1\pm \sqrt{25}}{4} $$ $$ t=\frac{-1\pm 5}{4} $$ Therefore $$ t_1=\frac{-1+ 5}{4}=\frac44=1 $$ $$ t_2=\frac{-1- 5}{4}=-\frac64=-\frac32 $$
$\endgroup$
