Question

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Answer 1

Question:-

If α, β are the zeroes of quadratic polynomial x² - p(x + 2) - c, then prove that (α + 2)(β + 2) - 4 + c = 0.

Answer:-

Given:-

α, β are the zeroes of quadratic polynomial x² - p(x + 2) - c = x² - px - 2p - c

On comparing the polynomial with standard form of a quadratic equation i.e., ax² + bx + c = 0 ;

Let;

• a = 1
• b = - p
• c = - 2p - c

We know that,

Sum of zeroes = - b/a

⟹ α + β = - ( - p)/1

⟹ α + β = p -- equation (1)

Product of zeroes = c/a

⟹ αβ = - 2p - c -- equation (2)

We have to prove:-

⟹ (α + 2)(β + 2) - 4 + c = 0

⟹ α(β + 2) + 2(β + 2) - 4 + c = 0

⟹ αβ + 2α + 2β + 4 - 4 + c = 0

⟹ αβ + 2(α + β) + c = 0

Putting the respective values from equations (1) , (2) we get,

⟹ - 2p - c + 2(p) + c = 0

⟹ - 2p + 2p = 0

⟹ 0 = 0

Hence, Proved.

Answer 2

Required Answer :-

We know that

$\bf Sum \; of \; zeroes = \dfrac{-b}{a}$

Where

b = -p

a = 1

$\sf Sum \; of \; zeroes = \dfrac{-(-p)}{1}$

$\sf Sum \; of \; zeroes = \dfrac{p}{1}$

$\sf Sum \; of \; zeroes = p$

Now

$\bf Product \; of \; zeroes = \dfrac{c}{a}$

Where

c = -2p -c

a = 1

$\sf Product \; of \; zeroes = \dfrac{-2p -c}{1}$

$\sf Product \; of \; zeroes = -2p-c$

Now

$\sf (\alpha + 2)\times(\beta + 2)-4+c=0$

$\sf (\alpha \times \beta ) + (\alpha \times 2) + (\beta \times 2) + (2 \times 2) - 4+c=0$

$\sf \alpha \beta +2\alpha +2\beta +4-4+c=0$

$\sf \alpha \beta +2\alpha +2\beta +c = 0$

Taking 2 as common

$\sf \alpha \beta + 2(\alpha +\beta )+c=0$

$\sf -2p-c+2\times p + c = 0$

$\sf-2p-c+2p+c=0$

$\sf -c + c=0$

$\sf 0=0$