another maths question...?

Question

Given that the equation
k x^2 - 4x + (k - 3 ) = 0
has real roots, show that
k^2 - 3k - 4 < 0.

Find that range of values of k satisfying this inequality.

thank you very much.

Answer 1

For the equation to have real roots, the discriminant must be non-negative.
4^2 - 4k(k - 3) >= 0
Dividing by 4:
4 - k(k - 3) >= 0
4 - k^2 + 3k >= 0
Multiplying by -1 and reversing the inequality sign:
k^2 - 3k - 4 <= 0.

Solving this equation:
(k - 4)(k + 1) <= 0

Either:

k - 4 <= 0 and k + 1 >= 0
k <= 4 and k >= -1
-1 <= k <= 4

or
k - 4 >= 0 and k + 1 <= 0
k >= 4 and k <= -1
This is not possible.

Therefore -1 <= k <= 4.

Answer 2

Use the quadratic formula on the first equation to get
x = (4 +/- sqrt(16 - 4k^2 + 12k)) / 2k

If x is real we must have 16 - 4k^2 + 12k > 0 (but see below).

-4k^2 + 12k + 16 > 0 (just rearranged)
4k^2 - 12k - 16 < 0 (when you multiply by -1 you have to reverse the inequality)
k^2 - 3k - 4 < 0 (dividing by 4 does not affect inequality)

(This problem is not quite correctly stated. The condition for x to be real is actually 16 - 4k^2 + 12k >= 0, which leads to the conclusion k^2 - 3k - 4 <= 0.)

To find the range of values satisfying this inequality, solve k^2 - 3k - 4 = 0.
We find k = (3 +/- sqrt(9 +16)) / 2 = 4, -1.

The polynomial in k describes a parabola that cuts the x-axis at -1 and 4 and thus has value less than 0 between those points. So the range of values satisfying the inequality is (-1, 4).

(If you allow k = 0, as you should, the range is [-1, 4], i.e. includes -1 and 4.)

Answer 3

1) When you use the quadratic equation and the part under the square root sign ( b^2 - 4ac ) is negative, you get imaginary roots. But we want real roots, not imaginary roots, so we want to find out when ( b^2 - 4ac) > 0. So your first step is using kx^2 - 4x + (k-3) = 0 and the quadratic equation to show that b^2 - 4ac > 0 only when k^2 - 3k - 4 < 0.

2) Then factor k^2 - 3k - 4 < 0, and find out the range of k values that will produce negative values.

Good luck!

Answer 4

the equation has two real roots

then discriminant > 0
b^2 - 4ac > 0

( -4 )^2 -4 ( k ) ( k-3 ) > 0
16 -4k^2 +12k > 0

-4k^2 + 12k + 16 > 0 divide by -4

k^2 - 3k - 4 < 0

Answer 5

ax^2 + bx + c = 0 has real roots if and only if:
b^2 - 4ac >0

So here: (-4)^2 - 4(k)(k-3)>0
16 - 4(k^2 - 3k)>0
16 > 4(k^2-3k) divide by 4 and take to the right
0 > k^2 -3k -4

k^2 - 3k - 4 = 0 => (k-4)(k+1)=0
And this is negative for -1

Answer 6

16-4k(k-3)>=0 (with = there is a real double root)
-4k^2+12k+16 >=0 so
k^2-3k-4<=0
k^2-3k-4=0
k=((3+-sqrt(25))/2
k= 4 and k= -1 the range is -1<=k<=4