<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>
There are no values of
m
m
that make the equation true.
No solution
Answer:
Step-by-step explanation:
You are multiplying the identity matrix by another matrix.
The effect of multiplying
1 0 0
0 1 0
0 0 1
by another matrix of the same dimensions is nil...nothing happens. So, the answer to this problem is the first matrix itself. Just copy it down.
20+x>70
-20. -20. on both sides to even it out
x>50 is the answer
