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Show that the equation $\displaystyle x^2+(3a-2)x+a(a-1)=0$ has real roots for all values of a are real numbers ..
I tried this ..
$\displaystyle (3a-2)^2-4(a^2-a)$
$\displaystyle =9a^2-12a+4-4a^2+4a$
$\displaystyle =5a^2-8a+4$
Is it because $\displaystyle 5a^2-8a+4$ is a quadratic equation and hence it has real roots ? Or i am in the wrong path ...
Hello,
In order to prove that the first equation (x²+(3a-2)x+a(a-1)=0) has real roots, you have to prove that the discriminant, 5a²-8a+4 is always positive (or equal to 0).Quote:
Originally Posted by mathaddict
So you have to see when $\displaystyle 5a^2-8a+4\geq 0$
In order to prove this, find the discriminant : 64-16*5<0
Hence the sign of 5a²-8a+4 is always the same, and its sign is the sign of the leading coefficient, 5.
Thus $\displaystyle 5a^2-8a+4>0$ for all a.
Therefore, the equation has real roots for all a.
41:):)th
you would have to show that the descriminant is always greater or equal zero for a as real number, so you would have to show that 5a^2-8a+4 is always greater than or equal to zero. You could do this by plotting the function on a graph.
sorry for repeating what was just said i think i posted it at the same time.
