If m and n are rhe roots of the quadratic equation 3x^2=6x+5 then find the value of (m+2n) (2m+n) and 1/m+1/n
Answers
Answer:
Step-by-step explanation:
m n are roots
then, m+n=-b/a=6/3=2
m*n=-5/3
m+2n*2m+n
=(m+n+n)(m+m+n)
=2*(m+n)^2+mn
=2*(2^2)-5/3
=19/3
1/m+1/n
=m+n/mn
=2/(-5/3)
=-6/5
Answer:
19/3, - 6/5
Step-by-step explanation:
Given, m and n are the roots of 3x² = 6x + 5 which simplifies, 3x² - 6x - 5 = 0
Let's take this in a objective point of view. So aim is to, Find the answer without finding roots.
We know that,
m + n = - b / a = 6/3 = 2
mn = - 5/3
Now,
Value of (i) (m+2n)(2m+n)
= (2m² + mn + 4mn + 2n²)
= (2m²+2n² + 4mn) + mn
=2 ( m² + n² + 2mn) +mn
= 2 ( m + n)² + mn
= 2 ( 2)² + (-5/3)
= 8 - 5/3
= 24 - 5 /3
= 19/3
Value of (ii) (1/m + 1/n)
= m + n / mn
= 2 / - 5/3
= 2/1 * 3/-5
= - 6/5