



 
Hi Claire, I expanded (x  k)(13x) + 1 = 0 to obtain a quadratic in x. I then used the general quadratic to solve for x and examined the discriminant, b^{2}  4ab. The discriminant is a quadratic in k. The quadratic equation (x  k)(13x) + 1 = 0 has distinct real roots of this discriminant is positive. Try this and see if you can show that the discriminant is positive. If you have trouble write back and tell us what you got for the discriminant and we will try to help. Harley Weston Claire wrote back. Hi Harley Here are the steps that I managed before getting stuck. (xk)(13k)+1=0 If the roots are real and distinct then b^{2}4ac > 0 (3k+1)^{2}  4(3)(k1) I can't factorise it so how do I show that it is positive. Please help. Thank you. Regards Claire, That's exactly what I got but you should say
I agree 9k^{2}6k+13 cant be factored so I solved 9k^{2}6k+13 = 0 using the general quadratic again. This time the discriminant is negative so the roots are not real! Consider the function
Its graph in the kzplane never crosses the kaxis since 9k^{2}6k+13 = 0 has no real solution. When k = 0 the value of z is 13 which is positive. Hence 9k^{2}6k+13 is positive for some value of k and to become negative it would have to cross the kaxis, which it doesn't. Hence 9k^{2}6k+13 is positive for all values of k. The graph of z = 9k^{2}6k+13 looks something like
Harley  


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