Question

Obtain a quadratic polynomial whose zeroes are 2alpha + betha and apha+ 2betha where apha and betha are the zeroes of xsquare +3x-10.

Answer 1

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Answer 2

==> In the example,

p(x) = x² + 3x - 10

a = 1
b = 3
c = -10

Let, α and β are the zeroes of the polynomial .

We know that,

α + β = -b/a

= -3/1

= -3

Also,

αβ = c/a

= -10/1

= -10

But, we have given that 2α + β and α + 2β are the zeroes of the given quadratic polynomial .

Sum According to these zeroes ,

α + β = 2α + β + α + 2β

= 2α + α + β + 2β

= 3α + 3β

= 3(a+ β)

= 3(-3) [ ∵ α+β = -3]

= -9

Product according to these zeroes

αβ = 2α + β * α + 2β

= 2α² + 4αβ + βα + 2β²

= 2α² + 5αβ + 2β²

= 2α² + 2β² + 5αβ

= 2(α² + β²) + 5(-10) [∵ αβ = -10]

= 2(α + β)² - 2αβ + (-50)

= 2(-3)² - 2(-10) -50 [∵ α+β= -3, αβ= -10]

= 2*9 +20 - 50

= 18 + 20 - 50

= 38 - 50

= -12

Now, we have

α + β = -9

and

αβ = -12


∴ The required quadratic polynomial is

=> x² -( α+ β)x + αβ

=> x² - (-9)x + (-12)

=> x² + 9x - 12


Thanks for your question!!!!

[ Be Brainly ]