Question

If α and β are the zeroes of the quadratic polynomial such that α + β = 24 and α – β = 8, find a quadratic polynomial having α and β as its zeroes.

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Answer 1

Solution:

We have,

α + β = 24 …… E-1

α – β = 8 …. E-2

By solving the above two equations accordingly, we will get

2α = 32 α = 16

Substitute the value of α, in any of the equation. Let we substitute it in E-2, we will get

β = 16 – 8 β = 8

Now,

Sum of the zeroes of the new polynomial = α + β = 16 + 8 = 24

Product of the zeroes = αβ = 16 × 8 = 128

Then, the quadratic polynomial is-K

x2– (sum of the zeroes)x + (product of the zeroes) = x2 – 24x + 128

Hence, the required quadratic polynomial is f(x) = x2 + 24x + 128

Answer 2

Explanation:

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α + β=24---->1

α - β=8------->2

Solving 1 and 2

2α =32

α=32÷2

α=16

And replacing in any equation

α - β=8

16-β=8

β=8

Now,...

Sum of roots=24

Product of roots=128

So,...

X²-sum of roots× X + product of roots=0

X²-24X +128=0

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