if alpha +beta are the roots of equation 4x²-5x+2=0 find the equation whose roots are1)alpha+3beta&3alpha+beta 2)alpha/betaβ/alpha 3)alpha²/betaβ²/alpha 4)alpha+1/alphaβ+1/beta
Answers
Answer:
16x² - 80x + 107 = 0
8x² - 9x + 8 = 0
4x² -5x + 4 = 0
4x² -15x + 23 = 0
Step-by-step explanation:
4x²-5x+2=0
α + β = 5/4
αβ = 2/4 = 1/2
Case 1 -
Roots α + 3β , 3α + β
(x -(α + 3β) ( x - (3α + β) = 0
=> x² - x (4α + 4β) + ( 3α² + 3β² + 10αβ) = 0
=> x² -4x(α+β) + 3(α+β)² + 4αβ = 0
=> x² -4x(5/4) + 3×(5/4)² + 4×(1/2) = 0
=> x² - 5x + 75/16 + 2 = 0
=> 16x² - 80x + 75 + 32 = 0
=> 16x² - 80x + 107 = 0
Case 2 -
Roots α/β , β/α
(x - α/β)(x -β/α) = 0
=> x² -x(α/β + β/α) + 1 = 0
=> x² - x(α² + β²)/αβ + 1 = 0
=> x² - x( (α + β)²-2αβ)/αβ + 1 = 0
=> x² - x( (5/4)² -2×(1/2))/(1/2) + 1 = 0
=> x² - 2x(25/16 -1) + 1 = 0
=> x² -2x(9/16) + 1 = 0
=> x² - 9x/8 + 1 = 0
=> 8x² - 9x + 8 = 0
Case 3
roots are α²/β , β²/α
(x-α²/β)(x - β²/α) = 0
=> x² -x(α²/β + β²/α) + αβ = 0
=> x² -x(α³ + β³)/αβ + αβ = 0
=> x² - x((α+β)³ -3αβ(α+β))/αβ + αβ = 0
=> x² - x((5/4)³ -3×(1/2)×(5/4))/(1/2) + 1/2 = 0
=> x² -2x ( 125/64 - 15/8) + 1/2 = 0
=> x² -2x (125 - 120)/8 + 1/2 = 0
=> x² - 5x/4 + 1/2 = 0
=> 4x² -5x + 4 = 0
Case 4
Roots are α + 1/α & β + 1/β
(x - (α + 1/α)( x -(β + 1/β) = 0
=> x² - x(α + β + 1/α + 1/β) + αβ + 1/αβ + α/β + β/α = 0
=>x² - x (5/4 + (β+α)/αβ) + 5/4 + 1/(1/2) + (β+α)/αβ = 0
=> x²- x(5/4 + (5/4)/(1/2) ) + 5/4 + 2 + (5/4)/(1/2) = 0
=> x² - x ( 5/4 + 5/2) + 13/4 + 5/2 = 0
=> x² - 15x/4 + 23/4 = 0
=> 4x² -15x + 23 = 0