Find the equation whose roots are:-3,2/3,1/2
Answers
Answer:
Here, S
1
=0+1+2+(−3)=0,
S
2
=0.1+1.2+2.(−3)+(−3).0+1.(−3)+0.2=−7
and S
3
=0.1.2+1.2.(−3)+2.(−3).0=−6,S
4
=0.1.2.(−3)=0
Thus required bi-quadratic is, x
4
−S
1
x
3
+S
2
x
2
−S
3
x+S
4
=0⇒x
4
−7x
2
+6x=0
Answer:
6x³ + 11x² - 19x + 6 = 0
Step-by-step explanation:
Given :- The given roots are - 3, 2/3 and 1/2.
To Find :- The equation whose roots are given.
Solution :-
The given roots are - 3 , 2/3 , 1/2 .
Since there are three roots, therefore it is a cubic equation.
A cubic polynomial equation is an equation whose variable has a highest degree of 3.
The cubic equation will be :-
( x - ( - 3 ) )( x - ( 2/3 ) )( x - ( 1/2 ) ) = 0
⇒ ( x + 3 )( ( 3x - 2 )/3 )( ( 2x - 1 )/2 ) = 0
⇒ (x + 3)( 3x - 2 )( 2x - 1 ) = 0
⇒ ( 3x² - 2x + 9x - 6 )( 2x - 1 ) = 0
⇒ 6x³ - 4x² + 18x² - 12x - 3x² + 2x - 9x + 6 = 0
⇒ 6x³ + 11x² - 19x + 6 = 0
Therefore, the equation whose roots are - 3, 2/3 and 1/2 is 6x³ + 11x² - 19x + 6 = 0
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