If the roots of 2X^3-X^2-22X-24=0 are in the ratio 3:4 then find the roots
Answers
Answer:
-3/2,and -2,4
Step-by-step explanation:
3a+4a+b=-(-24)/2=1/2
12a^2+4ab+3ab=-22/2=-11
3a(4a)b=-(-24)/2=12
On simplification we have
7a+b=1/2,12a^2+7ab=-11 and 12a^2b=12.
Eliminating b we get
22a^2+9a-1=0=>a=1/11,-1/2 and b=-3/22,4
Your roots would be -3/2,-2,4
Step-by-step explanation:
Given :-
The roots of the equation 2x^3-x^2 -22x -24 = 0 are in the ratio 3:4
To find :-
Find the roots ?
Solution :-
Given equation is 2x^3-x^2 -22x -24 = 0
On comparing with the standard cubic equation ax^3+bx^2+cx+d= 0
a = 2
b = -1
c = -22
d = -24
Ratio of the roots of the equation = 3:4
Let they be 3a and 4a
We know that
Sum of the roots = -b/a
=> 3a + 4a = -(-1)/2
=> 7a = 1/2
=> a = (1/2)/7
=> a = 1/(2×7)
=> a = 1/14
Therefore, Value of a = 1/14
The roots are 3a = 3(1/14) = 3/14
and 4a = 4(1/14) = 4/14 = 2/7
Answer:-
The roots are 3/14 and 2/7
Used formulae:-
→ The standard cubic equation ax^3+bx^2+cx+d= 0
→ Sum of the roots = -b/a