Question

if a and B are the zeroes of quadratic polynomial x2-5 then form a quadratic polynomial whose zeroes are 1+alpha and 1+ beta .​

Answer 1

Answer:

GM friend

Step-by-step explanation:

quadratic polynomial whose zeros are 1 + α and 1 + β will be x² - 2x  - 4

Given ,

x² - 5 = 0

=> (x-√5)(x + √5) = 0

=> x = ±√5

hence

α = √5 and β = -√5

new roots

= 1 + α  and 1  + β

= 1 + √5  and 1  - √5

hence the new equation will be

[x-(1+α)][x-(1+β)] = 0

=> [x-(1+√5)][x-(1-√5)] = 0

=> x² -(1+√5)x - (1-√5)x + (1+√5)(1-√5) = 0

=> x² - 2x + 1 - 5 = 0

=> x² - 2x  - 4 = 0

which is the required quadratic equation.

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Answer 2

Answer:

${x}^{2} - 5 = 0 \\ a = 1 \: \: \: \\ b = 0 \\ c = - 5 \\ d = 2 \sqrt{5} \\ x = + - 2 \sqrt{5} \div 2 \\ \alpha = + \sqrt{5} \\ \beta = - \sqrt{5} \\ (1 + \alpha )(1 + \beta ) = (1 + \sqrt{5} )(1 - \sqrt{5} ) \\ = - 4 \\ quadratic \: poynomial = (x + 4)(x + 4) \\ = {x}^{2} + 8x + 16$

Step-by-step explanation:

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