<span>Let p, np be the roots of the given QE.So p+np = -b/a, and np^2 = c/aOr (n+1)p = -b/a or p = -b/a(n+1)So n[-b/a(n+1)]2 = c/aor nb2/a(n+1)2 = cor nb2 = ac(n+1)2
Which will give can^2 + (2ac-b^2)n + ac = 0, which is the required condition.</span>
Answer:
18
Step-by-step explanation:
MJ IS 18 RHIS IS YOUR CORRECT ANSWER
I mean... in order to try each one you need to try it atleast once so.. 1 might be your answer
The -3 would be where your line crosses the y-axis, so you would plot your point at -3. your slope will always be counted as rise/run, basically meaning you would count along the y-axis a certain amount of times, in this case -1, and along the x-axis a certain amount of times, here would be three. your next plot point would be (3,-4) and so on. hope this helped!
