Ok, the question is:
"Given that the equation x^2 + 4x + c =0 has unequal real roots, find the range of possible values for c"
You are given that x^2 + 4x + c = (x+2)^2 -4 +c
The way I would do this question, is look at the equation x^2 + 4x + c =0, so a=1, b=4 and c=c. Substitute that into b^2-4ac>0 to find the answer, because b^2-4ac>0 when there are unequal real roots.
I would do this, getting 16-4ac>0, which then simplifies to 4>c.
the mark scheme has this method of doing it:
Ok, so they add 4 and subtract c to both sides in order to get (x+2)^2 = 4-c. After that, they then get to 4-c>0. How on earth do they get to this step from the previous step? Can you please explain?!
I can see that 4-c is on the RHS of the equation, but why is that then an answer for the discriminant. Without substituting a=1, b=4 into b^2-4ac to find the discriminant, how do they suddenly get to the stage where they have 4-c>0 from that equation?
I get that the discrimant, b^2-4ac, must be >0. But how do they get the value of the discriminant to be 4-c by using the previous step of (x+2)^2 = 4-c?
Can you explain please?
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- Thread Starter
Last edited by blobbybill; 06-11-2016 at 19:11.
- 06-11-2016 19:10
- 06-11-2016 19:16
If 4-c<0, then if you square root both sides of the equation what do you get? Can you find the square root of a negative number? If 4-c<0 then this implies that which should make you feel a bit uncomfortable, think to yourself of a number that when you square it you get a negative result. Can you think of such a number?