Question

If alpha and beta are the roots of the quadratic equation x^2 - p(x + 1) - c = 0, then (alpha + 1) (beta +1)

Answer 1

$\large\bf\underline{Given:-}$

• p(x) = x² -p(x+1)-c
• α and β are the roots of the given quadratic equation.

$\large\bf\underline {To \: find:-}$

• value of (α + 1) (β + 1)

$\huge\bf\underline{Solution:-}$

$\dashrightarrow \rm \: {x}^{2} - p(x + 1) - c \\ \\ \dashrightarrow \rm {x}^{2} - px - p - c \\ \\ \dashrightarrow \rm {x}^{2} - px - (p + c)$

So, p(x) = x² - px -(p+c)

where,

• a = 1
• b = -p
• c = -(p+c)

$\longrightarrow \bf \bigstar \: \: \alpha + \beta = \frac{ - b}{a} \\ \\ \longrightarrow \rm \alpha + \beta = \frac{ - ( - p)}{1} \\ \\ \longrightarrow \rm \alpha + \beta = p.........(i)$

$\longrightarrow \bigstar \: \bf\alpha \beta = \frac{c}{a} \\ \\ \longrightarrow \rm \alpha \beta = \frac{ - (p + c)}{1} \\ \\ \longrightarrow \rm \alpha \beta = - (p + c).......(ii)$

Now, finding value of (α + 1)(β + 1)

$: \implies \rm \: ( \alpha + 1)( \beta + 1) \\ \\ : \implies \rm \: \alpha ( \beta + 1) + 1( \beta + 1) \\ \\ : \implies \rm \: \alpha \beta + \alpha + \beta + 1 \\ \\ : \implies \rm \: ( \alpha + \beta ) + ( \alpha \beta ) + 1 \\ \\ \dag \: \rm \: substituting \: value \: of \: (\alpha + \beta ) \: \\ \rm \: \: \: \: and \: \alpha \beta \: from \: eq.(i) \: and \: (ii). \\ \\ : \implies \rm \: ( \alpha + \beta ) + ( \alpha \beta ) + 1 \\ \\ : \implies \rm \:p + \{ - (p + c) \} + 1 \\ \\ : \implies \rm \: \cancel{p} - \cancel{ p} - c + 1 \\ \\ : \implies \rm \:1 - c$

So, Value of (α + 1)(β + 1) = 1 - c

Answer 2

Given quadratic equation x² - p(x+1) - c = 0

⇒ x² - px - p - c = 0

⇒ x² - px - ( p + c ) = 0

Now compare given equation x² - px - ( p + c ) = 0 with ax²+bx+c=0 , we get ,

• a = 1 , b = - p , c = - ( p + c )

Sum of roots , α + β = -b/a

⇒ α + β = -(-p)/1

⇒ α + β =  p __ (1)

Product of roots , αβ = c/a

⇒ αβ = - (p+c)/1

⇒ αβ = - (p+c) __ (2)

Let's come to our required ,

⇒ (α+1)(β+1)

⇒ α (β+1) + (β+1)

⇒ αβ + α + β + 1

⇒ (αβ) + (α+β) + 1

⇒ - (p+c) + (p) + 1 [ From (1) & (2) ]

⇒ 1 - c

So the value of (α+1)(β+1) is (1-c)