If α and β are zeroes of 2x² - 5x+7, find

polynomial whose zeroes are 2α + 3β , 2β + 3α .

α & β are the zeroes of 2x² - 5x + 7.

polynomial whose zeroes are 2α + 3β , 2β + 3α .

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.

d) the triangle is a right triangle 60 squared +144 squared= 156 squared

step-by-step explanation:

156 squared is the hypothenuse

so 156= c

step-by-step explan mxfcg'k; uafwgzfgjhklhgation: