Question

If a, b, c are the roots of x^3 + 2x^2 - 4x -3= 0 then the equation whose roots area/3,b/3,c/3 is
1) x2 +6x2 – 36x-81=0
2) 9x + 6x2 - 4x-1=0
3) 9r3 + 6x2 + 4x+1=0
4) x2 - 6x² + 36x +81=0...​

Answer 1

Answer:

GIVEN EQUATION :–

${x}^{3} + 2 {x}^{2} - 4x - 3 = 0$

IT'S ROOTS ARE a , b , c .

TO FIND :–

A NEW QUADRATIC EQUATION WHICH ROOTS ARE a/3 , b/3 , c/3 .

SOLUTION :–

$sum \: \: of \: \: roots \: \: = \alpha _{1} = \frac{ - (cofficient \: \: of \: \: {x}^{2} }{leading \: \: cofficient}$

$a + b + c = - 2$

SO, SUM OF ROOTS OF NEW EQUATION –

$\frac{a}{3} + \frac{b}{3} + \frac{c}{3} = \frac{a + b + c}{3} = - \frac{2}{3}$

$\alpha _{2} = ab + bc + ca = \frac{cofficient \: \:of \: \: x }{l.c. } \\ \\ ab + bc + ca = - 4$

SO , FOR NEW EQUATION –

$\frac{ab}{9} + \frac{bc}{9} + \frac{ca}{9} = - \frac{4}{9}$

$multiply \: \: of \: \: roots( \alpha _{3}) = abc = - \frac{constant}{l.c.} \\ \\ abc = 3$

FOR NEW EQUATION –

$\frac{abc}{3 \times 3 \times 3} = \frac{abc}{27} = \frac{1}{9}$

SO , NEW EQUATION –

${x}^{3} + \frac{2}{3} {x}^{2} - \frac{4}{9} x - \frac{1}{9} = 0$

$9 {x}^{3} + 6 {x}^{2} - 4x - 1 = 0$

OPTION (2) IS CORRECT...

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