1. Using analytical method, solve the inequality x^2 - 4x + 5
<0
I solved this question by completing the sqyare and got (x-2)^2
+ 1 <0
I was stuck after that and there was this "alternative method"
provided for us. It says," Since the coefficient of x^2 is positive
and discriminant, D = (-4)^2 - 4(5) < 0, therefore, teh graph of
y = x^2 - 4x + 5 is above the x-axis, i.e. x^2 - 4x +
5>0 for all real vlaues of x. Hence x^2 - 4x + 5 < 0
has no solution.
I don't get why x^2 - 4x + 5>0. As seen by
my solution using completing the square, I followed the initial
sign and still got a '<' sign. Anyway, what do i do after
completing the square?
2. Using an algebraic method, solve the inequality x^2 + 3x - 1
> 0 (solve this using completing the square, stated by
question)
I got (x + 1.5)^2 - 13/4 > 0
Subsequently, i do not know what steps to take to solve the
question (find the value of x)
Thanks
Hi anpanman,
For 1), notice that (x-2)^2 + 1 will always be more or equal to
1, since (x-2)^2 will always be more or equal to 0. Thus, the
inequality x^2 - 4x + 5 < 0 has no solution.
When you are using the "completing the square" method, you
should not bring the < 0 in for this case.
For 2), From (x+(3/2))^2 - 13/4 > 0, bring 13/4 to the other
side to get (x+(3/2))^2 > 13/4. Square root both side, and
you will realise that x + 3/2 must be less than [- (sqrt 13/4)] or
more than (sqrt 13/4). I believe you should be able to solve this
from here.
Cheers.