# For what values of k does kx2+12x+9=0 have imaginary roots?A detailed (step by step) answer would be great!

### 3 Answers | Add Yours

The function will have imaginary roots when the determinant is negative. The determinant = (b(squared)-4ac

so: 12(squared) - 4(k)(9)<0

144-36k<0

-36k<-144

k>4

In order to see why the roots of a second grade equation could be real or imaginary, let's see first how to find them.

First of all, the general form of a second grade equation is:

ax^2 + bx + c=0

Let's force the coefficient "a to be a common factor.

a*[x^2 + (b/a)*x +(c/a)]=0

Because "a" is different from zero, otherwise the equation would degenerate from the second grade to the first grade, we could divide the expression above with "a".

x^2 + (b/a)*x+ (c/a)=0

We'll try to create in the expression above , the development of the binomial (x+(b/2a))^2.

[x^2 + 2*(b/2a)*x + (b^2/4a^2)]- (b^2/4a^2)+(c/a)=0

The first 3 terms are the the development of the binomial (x+(b/2a))^2. Between the last 2 terms we'll find the common denominator, which is 4a^2. For this reason,we'll amplify the last term, (c/a), with the value (4a).

(x+(b/2a))^2 + (4ac - b^2)/4a^2=0

We'll move the free term to the right side of the equality and we'll keep the unknown to the left side.

(x+(b/2a))^2 = - (4ac - b^2)/4a^2

(x+(b/2a))^2 = (b^2-4ac)/4a^2

In order to find the unknown x, first we have to apply square root from the right term.

x+(b/2a) = sqrt [(b^2-4ac)/4a^2]

x=+/-[sqrt (b^2-4ac)]/2a - (b/2a)

Now, in order to find out the sqrt from (b^2-4ac)], the expression (b^2-4ac) has to be positive, because the square root is even.

**If (b^2-4ac)>=0, then the roots are real, but if (b^2-4ac) is negative, the roots are complex (imaginary). So, the expression (b^2-4ac) is the one to decide if the equation has ral or imaginary roots.**

In the case of the equation kx2+12x+9=0, the coefficients a=k, b=12 and c=9.

(b^2-4ac) = 144-4k*9=144-36k

In order to have complex roots, 144-36k<0

144<36k

12*12<4*3*3*k

12*12<12*3*k

12<3*k

3*4<3*k

**4<k**

kx^2+12x+9=0.

To determine the values of k for which the roots are imaginary.

Solution:

kx^2+12x+9=k(x^2+(12/k) x+9/k) = 0 or

k{x+(12/k)x+(6/k)^2 - (6/k)^2+9/k} = 0

k{x+6/k}^2 - (36-9k)/k^2 = 0 or

{x+6/k)^2 = (36-9k)/k = 9(4-k)k

If (4-k)/k is < 0, then the (x+6/k)^2 , a perfect square ,is -ve . This implies the roots are imaginary.

(4-k)/k <0 if 4/k <1 or k >4 for all k > 4.