Exam Question 2

(k+1)x^2+12x+(k-4)=0 , real roots

• a). Show: k^2-3k-40≤0
• b). Hence find the possible values of k

(k+1)x^2+12x+(k-4)=0 , real roots

• a). Here we will need to use the discriminant to show the equation is less than or equal to 0. However, we read that the original has real roots, so how can the equation be less than or equal to 0? Well, stick with it having real roots and let the maths do the talking. So b^2-4ac≥0 and now we sub in the values. a=(k+1) , b=12 , c=(k-4) , so 12^2-4(k+1)(k-4)≥0. Expanding this and collecting the terms gives 160-4k^2+12k≥0, which can be divided by 4 to give 40-k^2+3k≥0.Finally, to make this less than or equal to 0 we need to flip the signs (subtract 40-k^2+3k from both sides). So if we tidy up 40-k^2+3k, we get more simply the quadratic -k^2+3k+40, and flipping the signs gives us k^2-3k-40, so now we can say k^2-3k-40≤0. The question is complete.
• b). Now we just need to solve the quadratic and express its inequalities. So k^2-3k-40=0 is also equal to (k-8)(k+5)=0, so the solutions are k=8 or k=-5 (which is a boundary). Expressed as an inequality, this is: -5≤k≤8. The question is complete