Question 5:
Find the range of values of k if the quadratic equation 3(x2 – kx – 1) = k – k2 has two real and distinc roots.
Solution:
3(x2−kx−1)=k−k23x2−3kx−3−k+k2=03x2−3kx+k2−k−3=0a=3,b=−3k,c=k2−k−3In cases of two real and distinc roots,b2−4ac>0 is applied.(−3k)2−4(3)(k2−k−3)>09k2−12k2+12k+36>0−3k2+12k+36>0−k2+4k+12>0k2−4k−12<0(k+2)(k−6)<0k=−2,6
The range of values of k is −2<k<6.
Find the range of values of k if the quadratic equation 3(x2 – kx – 1) = k – k2 has two real and distinc roots.
Solution:
3(x2−kx−1)=k−k23x2−3kx−3−k+k2=03x2−3kx+k2−k−3=0a=3,b=−3k,c=k2−k−3In cases of two real and distinc roots,b2−4ac>0 is applied.(−3k)2−4(3)(k2−k−3)>09k2−12k2+12k+36>0−3k2+12k+36>0−k2+4k+12>0k2−4k−12<0(k+2)(k−6)<0k=−2,6

The range of values of k is −2<k<6.
Question 6:
Given that the quadratic equation hx2 – (h + 2)x – (h – 4) = 0 has real and distinc roots. Find the range of values of h.
Solution:
The quadratic equation hx2−(h+2)x−(h−4)=0has real and distinc roots.b2−4ac>0 is applied.(−h−2)2−4(h)(−h+4)>0h2+4h+4+4h2−16h>05h2−12h+4>0(5h−2)(h−2)>0The coefficient of h2 is positive, the region above the x-axis should be shaded.(5h−2)(h−2)=0h=25,2
The range of values of h for (5h−2)(h−2)>0 ish<25 or h>2.
Given that the quadratic equation hx2 – (h + 2)x – (h – 4) = 0 has real and distinc roots. Find the range of values of h.
Solution:
The quadratic equation hx2−(h+2)x−(h−4)=0has real and distinc roots.b2−4ac>0 is applied.(−h−2)2−4(h)(−h+4)>0h2+4h+4+4h2−16h>05h2−12h+4>0(5h−2)(h−2)>0The coefficient of h2 is positive, the region above the x-axis should be shaded.(5h−2)(h−2)=0h=25,2

The range of values of h for (5h−2)(h−2)>0 ish<25 or h>2.