both of the current solutions are incorrect. here's the lowdown:
number one
san Wrote:Question Regarding inequality. I have a doubt.
If it is given that z² - 4z > 5.
Can we write like this: z(z-4)>5
This implies either z>5 or z-4>5
z-4>5 implies z>9?
can't do that.
when you solve a
quadratic inequality - just as when you're solving a quadratic equation - you need to
get zero on one side, and THEN factor.
here's the reason why:
if (this)(that) = 0, then you know that either this = 0 or that = 0.
by contrast, if (this)(that) = 5, then you have no idea what either this or that is. (they could be 1 and 5; 3/2 and 10/3; -√3 and -5√3/3; etc)
same thing with inequalities. if you know (this)(that) < 0, then you know 'this' and 'that' have opposite signs. by contrast, if you know (this)(that) < 5, then you really don't know much at all.
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number two
solving a quadratic inequality
z² - 4z > 5
z² - 4z -5>0
(z-5)(z+1)>0
z>5, z>-1
good until the last step, but then you choked there.
first of all, "z > 5, z > -1" isn't a reasonable solution, because one of the two conditions becames completely irrelevant if you write it that way. (specifically, "z > 5 AND z > -1" is equivalent to just saying z > 5, and "z > 5 OR z > -1" is equivalent to just saying z > -1.)
here's what you do:
move everything over to get 0 on one side (check)
factor (check)
THEN, find the "boundary numbers", which are the numbers that would be the solutions if this were an
equation. in this case, those "boundary numbers" are -1 and 5.
once you find the boundary numbers,
set up a number line, and test the different regions into which the line is divided by the boundary numbers.
in this problem, your number line looks like this:
<--------------------(-1)----------------------(5)--------------------------->
there are 3 regions: x < -1, -1 < x < 5, and x > 5.
test each of the regions in one of two ways: either (a) plug in a number that lies in each of the regions and see whether the original inequality is true, or (b) just think about the signs of the factors in the factored version of the inequality.
if you're going to do method (a), you could plug in, say, -4 for the left hand region (for which the inequality would be true), 0 for the middle region (false), and 6 for the right hand region (true).
if you're going to do method (b):
left hand region --> (x - 5) and (x + 1) are both negative --> (x - 5)(x + 1) > 0. true
middle region --> (x - 5) is negative, (x + 1) is positive --> (x - 5)(x + 1) < 0. false
right hand region --> (x - 5) and (x + 1) are both positive --> (x - 5)(x + 1) > 0. true
either way, you get the correct solution, which is
x < -1 or x > 5.