Question

If alpha, ß are the roots of 3x² - 5x + 7 =0 then find (i) alpha square +beta square (ii) alpha cube + beta cube
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Answer 1

Answer:

P(x)= 3x^2-5x+7

If alpha and beta are the roots then,

Alpha + beta= -b/a= -(-5)/3= 5/3

Alpha × beta=c/a= 7/3

(i) Alpha ^2 + beta^2 = ?

(alpha + beta) sq.= alpha sq.+ beta sq.+ 2×alpha×beta

(5/3)^2= alpha sq.+ beta sq.+ 2×7/3

25/9= alpha sq.+ beta sq.+14/3

25/9 - 14/3= alpha sq. +beat sq.

(By taking LCM)

25/9 - 42/9= alpha sq. + beta sq.

Therefore,

alpha sq. + beta sq.= -17/9

(ii) Alpha^3 + beta^3 = ?

(alpha + beta)^3= alpha^3 + beta^3+ 3×alpha× beta( alpha+ beta )

(5/3)^3 = alpha^3 + beta ^3 + 3× 7/3 ×(5/3)

125/27 = alpha^3 + beta^3 + 7×5/3

125/27 = alpha^3 + beta^3 + 35/3

125/27 - 35/3 = alpha^3 + beta^3

(By taking LCM)

125/27 - 315/9 = alpha^3 + beta^3

Therefore,

Alpha^3 + beta^3 = -190/ 27

Answer 2

$\red{\large{\bf{QuEsTion}}}$

If α and β are the roots of 3x² - 5x + 7 = 0

then find

(i) α² + β²

(ii) α³ + β³

$\red{\large{\bf{SoluTion}}}$

As we know,

$\tt sum\:of\:roots\:=\frac{-(coefficient\:of\:x)}{coefficient\:of\:x^2}$

so,

α + β= -(-5)/3 = 5/3 ....eqn(1)

also as we know,

$\tt product\:of\:roots=\frac{constant \:term}{coefficient\:of\:x^2}$

therefore,

α β = 7 / 3 .......eqn(2)

$\blue{\bf{Solution\:(i)}}$

α² + β² = ?

using identity

( x²+y² = (x+y)²-2xy)

α² + β² = (α + β)² - 2 α β

(using eqn(1) and (2))

α² + β² = ( 5 / 3)² - 2(7/3)

            = (25 / 9) - (14/3)

            = ( 25 - 42 ) / 9

             = -17 / 9

$\boxed{ \alpha ^{2} +\beta ^{2} = \frac{-17}{9} }$

$\blue{\bf{Solution\:(ii) }}$

α³ + β³ = ?

(using identity

x³ + y³ = (x+y)³ - 3 x y(x+y) )

α³ + β³ = (α +β)³ - 3αβ(α+β)

(using eqn(1) and (2))

α³ + β³ = (5/3)³ - 3(7/3)(5/3)

           = (125/27) - (35/3)

           = -190 / 27

$\boxed{\alpha ^{3} +\beta ^3 =\frac{-190}{27} }$