<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:I dont know but i think it is 1/2
Step-by-step explanation:
hope this helps
-4^2 is the same as -4*-4
-4*-4=16
-4^2= 16
-4^2 is the exponent.
I hope this helps!
~kaikers
The remainder for 67 divided by 3 would be 1 because (this might be a bad explanation since i'm not good at explaining math without showing u my work) if u do divide 6 by 3, 3x2 is 6 and 6-6 is 0. Then if u divide 7 by 3, then 6 is the closest u can get to 7 without multiplying by a decimal. Then if u do 7-6, u get 1 and u cant divide 1 by 3 so the remainder would be 1.
