Question

If \[\alpha ,\beta \] are the zeroes of the polynomial \[f(x) = {x^2} - p(x + 1) - c = 0\] such that \[(\alpha + 1)(\beta + 1) = 0\], then \[c = \]
a) \[1\]
b) \[0\]
c) \[ - 1\]
d) \[2\]

Answer 1

Hint: In the given Polynomial, we will first separate the coefficients of \[{x^2},x\] and the constant terms. Now, we know, if \[\alpha ,\beta \] are the zeroes of the polynomial of the form \[u{x^2} + vx + w = 0\], then \[\alpha + \beta = \left( {\dfrac{{ - v}}{u}} \right)\] and \[\alpha \beta = \left( {\dfrac{w}{u}} \right)\]. We will then get some conditions on \[\alpha \] and \[\beta \]. Also, in the question, For \[\alpha \] and \[\beta \] we are given one condition to be satisfied. We will obtain the equations in terms of \[\alpha \] and \[\beta \] from the given condition and solve for the given polynomial. Then, we will solve the obtained equations to find the value of \[c\].

Complete step by step solution:
Now, we are given \[\alpha ,\beta \] are the zeroes of \[f(x) = {x^2} - p(x + 1) - c = 0\]
First of all, separating the coefficients of \[{x^2},x\] and the constant terms.
\[f(x) = {x^2} - p(x + 1) - c = 0\]
Opening brackets of the equation
\[f(x) = {x^2} - px - p - c = 0\]
Compiling constant terms together, we get
\[f(x) = {x^2} - px - (p + c) = 0\]
Now, comparing this polynomial with \[u{x^2} + vx + w = 0\], we get
Coefficient of \[{x^2}\] \[ \Rightarrow u = 1\]
Coefficient of \[x\] \[ \Rightarrow v = - p\]
Constant terms \[ \Rightarrow w = - (p + c)\]
Now, using \[\alpha + \beta = \left( {\dfrac{{ - v}}{u}} \right)\] and \[\alpha \beta = \left( {\dfrac{w}{u}} \right)\] , we get
\[\alpha + \beta = \left( {\dfrac{{ - ( - p)}}{1}} \right)\] and \[\alpha \beta = \left( {\dfrac{{ - (p + c)}}{1}} \right)\]
We know, \[( - ( - x) = x)\]. So,
\[\alpha + \beta = \left( {\dfrac{p}{1}} \right)\] and \[\alpha \beta = \left( {\dfrac{{ - (p + c)}}{1}} \right)\]
As \[\left( {\dfrac{x}{1} = x} \right)\], we can write the above expressions as :
\[\alpha + \beta = p\] and \[\alpha \beta = - (p + c)\]
Hence, we got
\[\alpha + \beta = p\] -----(1)
\[\alpha \beta = - (p + c)\] -----(2)
Also, we are given \[(\alpha + 1)(\beta + 1) = 0\]
Solving this, we get
Using \[(a + b)(c + d) = a(c + d) + b(c + d)\]
\[\alpha (\beta + 1) + 1(\beta + 1) = 0\]
Now, opening the brackets and solving
\[\alpha \beta + \alpha + \beta + 1 = 0\]
Hence, we have
\[\alpha \beta + \alpha + \beta + 1 = 0\] ------(3)
Now, putting the value of \[\alpha + \beta \] and \[\alpha \beta \] from (1) and (2) in (3), we get
\[ - (p + c) + p + 1 = 0\]
\[ - p - c + p + 1 = 0\]
As \[ - p + p = 0\], we are left with
\[ - c + 1 = 0\]
Now, adding \[c\]both the side, we get
\[ - c + 1 + c = 0 + c\]
We know, \[0 + x = x\], so the above equation becomes
\[1 = c\]
\[\therefore c = 1\]
Hence, we got \[c = 1\].
So, the correct answer is “Option a”.

Note: We need to take care where we have to take the negative sign and where positive while we are calculating the sum and product of zeroes of a polynomial. Also, we need to consider each and everything given in the question. We can’t just ignore the given conditions and solve for the other. We could have solved the equations by eliminating terms also, but that would become confusing and too lengthy so, we did this by this easy to solve method.