Question

If alpha and beta
are the zeroes of the
quadratic polynomial
f(x) = x^2 -p ( x+ 1) -c , show that
(α+ 1) (β + 1 ) = 1 - c​

Answer 1

To ProvE :

• (α+ 1) (β + 1 ) = 1 - c

SolutioN :

$\tt \dagger \: \: \: \: \: Sum \: of \: Zeros.$

$\tt \dagger \: \: \: \: \: \alpha + \beta = \dfrac{ - b}{a}$

$\tt \dagger \: \: \: \: \: product \: of \: Zeros.$

$\tt \dagger \: \: \: \: \: \alpha \beta = \dfrac{c}{a}$

A/Q,

We have Equation,

$\tt \dagger \: \: \: \: \: {x}^{2} - p(x + 1) - c.$

$\tt : \implies {x}^{2} - p(x + 1) - c.$

$\tt \dagger \: \: \: \: \: {x}^{2} - px - p - c.$

Compare with General Expression.

$\tt \dagger \: \: \: \: \: a {x}^{2} + bx + c$

Where as,

• a = 1.
• b = -p
• c = - p - c.

Now,

→ (α+ 1) (β + 1 )

→ αβ + α + β + 1.

• Product of Zeros ( αβ ) → - p - c
• Sum of Zeros ( α + β ) → p.

★ Putting the given value on it.

→ αβ + α + β + 1.

→ - p - c + p + 1.

→ - c + 1.

Hence Proved.