please answer this question..
Answers
a,b and c are real numbers. m is an integer.
a+b+c = 6 --- (1) ab+bc+ca = 9 --- (2)
Eq (2): ab = 9–(a+b)c = 9–(6-c)c = c² -6c+9
We know (a-b)² ≥ 0 for real a and b.
=> (a+b)²-4ab ≥ 0
=> (6 - c)² ≥ 4 (c² - 6c + 9)
=> c² - 12c +36 ≥ 4 c² - 24c + 36
=> 3 c² - 12 c ≤ 0 or c ( c - 4) ≤ 0
So 0 ≤ c ≤ 4
Let r1 and r2 be
the roots of x² - (m+2) x + 5 m = 0
=> Let r1 = [m+2 - √(m²-16m+4) ] /2 ,, r2 =[m+2
+ √(m²-16m+4)] /2
We see that clearly r1 ≤ r2.
m²-16m+4 ≥ 0, only if m≤0 or m≥16 (integers).
(we can know this by finding its roots).
For m ≥ 16, r1 > 4 and also r2 > 4. Not valid case.
So consider only m ≤ 0. Let m = -n. n ≥ 0.
=> r1 = [2-n - √(n²+16n+4)]/2, r2 = [2-n + √(n²+16n+4)]/2
We see clearly that r1 < 0 always . So 0 ≤ r2 ≤ 4
2 - n < √(n²+16n+4)
If 2-n < 0, then this condition is satisfied directly.
If 2-n ≥ 0, then n²-4n+4 < n²+16n+4
or n > 0
ie., m < 0
Now r2: 0 ≤
[2-n + √(n²+16n+4)]/2 ≤ 4
n-2 ≤ √(n²+16n+4)
≤
n+6
The left side condition is always satisfied.
Then n²+16n+4 ≤ n²+12n+36
=> 4 n ≤ 32 or n ≤ 8 or -8 ≤ m ≤ 0
So the integer values of m are [-8, -1].