SOLUTION: could someone assist me on how i can crack this problem it says in the quadratic equation Kx^(2) + 2(k+1)x + (k-1)=0 where k is a constant .solve the equation in the case K=5 .fi

Question 411549: could someone assist me on how i can crack this problem it says in the quadratic equation Kx^(2) + 2(k+1)x + (k-1)=0 where k is a constant
.solve the equation in the case K=5
.find the set of values for k which the equation has distinct real roots
thanks in advance!!
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
when k = 5, your expression becomes:

5x^2 + 12x + 4 = 0

A graph of this equation looks like this:

You will have real roots as long as the discriminant of b^2 - 4ac is greater than or equal to 0.

If it is negative, then you will have imaginary roots.

This comes from the quadratic formula of:

b^2 - 4ac, being under the square root sign, will not allow a real answer unless that value is greater than or equal to 0.

The general form of your equation is:

kx^2 + 2(k+1)x + (k-1)

set it equal to 0 and you have the standard form of the quadratic equation.

you get:

kx^2 + 2(k+1)x + (k-1) = 0

Since the standard form is:

ax^2 + bx + c = 0, then you get

a = k
b = 2*(k+1) = 2k+2
c = k-1

from this you get:

b^2 = (2k+2)^2 = 4k^2 + 8k + 4
4ac = 4 * k * (k-1) = 4k^2 - 4k

b^2 - 4ac = 4k^2 + 8k + 4 - (4k^2 - 4k) which leads to:

b^2 - 4ac = 4k^2 + 8k + 4 - 4k^2 + 4k which leads to:

b^2 - 4ac = 12k + 4

what you need is for the expression 12k + 4 to satisfy the equation:

12k + 4 >= 0

solving this, you get:

k >= (-1/3)

If we set k = -(1/3), then our equation would become:

(-1/3)x^2 + (4/3)*x - (4/3) = 0

The graph of this equation looks like this:

You can see that we have a real root at x = 2.

If we let k equal something less than -1/3, then we will not have real roots.

As an example, let k = -1.

The equation becomes -x^2 -2 = 0

A graph of this equation looks like this:

The graph doesn't cross the x-axis so there are no real roots.

I do not believe that k can be equal to 0, because they you don't have a quadratic equation anymore.

In that case you have a linear equation.

that equation would be 2x - 1

If we let k = - 1/4 which is close to but greater than -1/3, then we should see 2 real roots.

When k = -1/4, the equation becomes:

(-1/4)x^2 + (6/4)x - (5/4)

The graph of that equation looks like this:

We can try another posiive value of k.

Choose k = 2.

Our equation becomes:

2x^2 + 6x + 1 = 0

the graph of that equation looks like this:

We could try more, but the basic fact of using the discriminant to determine when you have real roots and when you don't has already determined the ranges in the value of k that would make the result have real roots.

You have real roots when k >= (-1/3).

You have imaginary roots when k < (-1/3).

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