Given:
α & β are the zeroes of 2x² - 5x + 7.
Find:
polynomial whose zeroes are 2α + 3β , 2β + 3α .
Solution:
On comparing with the standard form of a quadratic equation,
Let,
a = 2
b = - 5
c = 7.
We know that,
Sum of the zeroes = - b/a
⟹ α + β = - ( - 5)/2
⟹ α + β = 5/2 -- equation (1)
Product of the zeroes = c/a
⟹ αβ = 7/2 -- equation (2)
Now,
General form of a Quadratic equation is x² - (sum of the zeroes)x + product of the zeroes = 0.
★ Sum of the zeroes = 2α + 3β + 2β + 3α
⟹ Sum of the zeroes = 5α + 5β
⟹ Sum of the zeroes = 5(α + β)
Putting the value from equation (1).
⟹ Sum of the zeroes = 5(5/2)
⟹ Sum of the zeroes = 25/2
Similarly,
★ Product of the zeroes = (2α + 3β)*(2β + 3α)
⟹ Product of the zeroes = 2α( 2β + 3α) + 3β( 2β + 3α )
⟹ Product of the zeroes = 4αβ + 6α² + 6β² + 9αβ
⟹ Product of the zeroes = 13αβ + 6(α² + β²)
a² + b² = (a + b)² - 2ab → α² + β² = (α + β)² - 2αβ
⟹ Product of the zeroes = 13αβ + 6[ (α + β)² - 2αβ ]
Putting the values from equations (1) & (2) we get,
⟹ Product of the zeroes = 13(7/2) + 6 [ (5/2)² - 2(7/2) ]
⟹ Product of the zeroes = 91/2 + 6 ( 25/4 - 14/2 )
Taking LCM we get,
⟹ Product of the zeroes = ( 2 * 91 + 6 * 25 + 2 * 14 ) /2
⟹ Product of the zeroes = (182 + 150 + 28 )/2
⟹ Product of the zeroes = 180
Hence,
New quadratic equation = x² - (25/2) * x + 180 = 0
⟹ New quadratic equation = (2x² - 25x + 360)/2 = 0
⟹ New quadratic equation = 2x² - 25x + 360 = 0
∴ The required quadratic equation is 2x² - 25x + 360 = 0.
I hope it will help you.
Regards.