Question

Refer to the Attachment.

Happy Answering.

Topic- Quadratic Equation (class 11)​

Answer 1

Answer in the attachment

Answer 2

QUESTION:

If α, β and γ are the roots of equation x³ - px + q = 0, then the cubic equation whose roots are [α/(1+α)], [β/(1+β)] and [γ/(1+γ)] will be?

SOLUTION:

Given:

• α, β and γ are the roots of equation x³ - px + q = 0

To Find:

• Cubic equation whose roots are [α/(1+α)], [β/(1+β)] and [γ/(1+γ)].

Answer:

• Option A : (p + q - 1)x³ - (2p + 3q)x² + (p + 3q)x - q = 0 is correct.

Step By Step Explanation:

We know that for a cubic equation, ax³ + bx² + cx + d = 0,

• Sum of zeroes = -b/a
• Sum of product of zeroes taken two at a time = c/a
• Product of zeroes = -d/a

So,

• Sum of zeroes = α + β + γ = -b/a = -(0)/1 = 0 ————(1)
• Sum of product of zeroes taken two at a time = αβ + βγ + γα = c/a = (-p)/1 = -p ————(2)
• Product of zeroes = αβγ = -d/a = -(q)/1 = -q ————(3)

Now, we need to find the cubic equation whose roots are [α/(1+α)], [β/(1+β)] and [γ/(1+γ)].

For this, we will first find the sum of zeroes, sum of product of zeroes taken two at a time and product of zeroes for the required cubic equation.

First, we will find sum of zeroes.

So,

$\implies\text{Sum of zeroes =} \dfrac{\alpha}{1+\alpha}+\dfrac{\beta}{1+\beta}+\dfrac{\gamma}{1+\gamma}$

$\implies\text{Sum of zeroes =}\dfrac{\alpha(1+\beta)(1+\gamma)+ \beta(1+\alpha)(1+\gamma)+ \gamma(1+\beta)(1+\alpha)}{(1+\alpha)(1+\beta)(1+\gamma)}$

$\implies\text{Sum of zeroes =}\dfrac{(\alpha+\alpha\gamma+\beta\alpha+\alpha\beta\gamma)+(\beta+\beta\gamma+\beta\alpha+\alpha\beta\gamma)+(\gamma+\alpha\gamma+\beta\gamma+\alpha\beta\gamma)}{1+\alpha+\beta+\alpha\beta+\gamma+\alpha\gamma+\beta\gamma+\alpha\beta\gamma}$

Grouping the terms,

$\implies\text{Sum of zeroes =}\dfrac{(\alpha+\beta+\gamma)+2(\alpha\beta+\beta\gamma+\gamma\alpha)+3(\alpha\beta\gamma)}{1+(\alpha+\beta+\gamma)+(\alpha\beta+\alpha\gamma+\beta\gamma)+(\alpha\beta\gamma)}$

Using (1), (2) and (3),

$\implies\text{Sum of zeroes =}\dfrac{(0)+2(-p)+3(-q)}{1+(0)+(-p)+(-q)}$

$\implies\text{Sum of zeroes =}\dfrac{-2p-3q}{1-p-q}=\dfrac{2p+3q}{p+q-1}$

Hence,

$\implies\text{Sum of zeroes =}\dfrac{2p+3q}{p+q-1}- - - -(4)$

Now, we will find sum of product of zeroes taken two at a time.

So,

$\implies\text{Sum of product of zeroes taken two at a time =} \left(\dfrac{\alpha}{1+\alpha}\times\dfrac{\beta}{1+\beta}\right)+ \left(\dfrac{\alpha}{1+\alpha}\times\dfrac{\gamma}{1+\gamma}\right)+ \left(\dfrac{\gamma}{1+\gamma}\times\dfrac{\beta}{1+\beta}\right)$

$\implies\text{Sum of product of zeroes taken two at a time =}\dfrac{\alpha\beta}{(1+\alpha)(1+\beta)}+ \dfrac{\alpha\gamma}{(1+\alpha)(1+\gamma)} + \dfrac{\gamma\beta}{(1+\gamma)(1+\beta)}$

$\implies\text{Sum of product of zeroes taken two at a time =}\dfrac{\alpha\beta(1+\gamma)+\alpha\gamma(1+\beta)+ \gamma\beta(1+\alpha)}{(1+\alpha)(1+\beta)(1+\gamma)}$

$\implies\text{Sum of product of zeroes taken two at a time =}\dfrac{\alpha\beta+\alpha\beta\gamma+ \alpha\gamma+\alpha\beta\gamma+ \gamma\beta+\alpha\beta\gamma}{1-p-q}$

We had calculated the value of (1+α)(1+β)(1+γ) to be (1-p-q), that’s why we put its value directly.

Grouping the terms,

$\implies\text{Sum of product of zeroes taken two at a time =}\dfrac{(\alpha\beta+\alpha\gamma+\gamma\beta)+3(\alpha\beta\gamma)}{1-p-q}$

Using (2) and (3),

$\implies\text{Sum of product of zeroes taken two at a time =}\dfrac{(-p)+3(-q)}{1-p-q}$

$\implies\text{Sum of product of zeroes taken two at a time =}\dfrac{-p-3q}{1-p-q}=\dfrac{p+3q}{p+q-1}$

Hence,

$\implies\text{Sum of product of zeroes taken two at a time =}\dfrac{p+3q}{p+q-1} - - - -(5)$

Now, we will find product of zeroes.

So,

$\implies\text{Product of zeroes =} \dfrac{\alpha}{1+\alpha}\times\dfrac{\beta}{1+\beta}\times\dfrac{\gamma}{1+\gamma}$

$\implies\text{Product of zeroes =} \dfrac{\alpha\beta\gamma}{(1+\alpha)(1+\beta)(1+\gamma)}$

$\implies\text{Product of zeroes =} \dfrac{\alpha\beta\gamma}{1-p-q}$

Using (3),

$\implies\text{Product of zeroes =} \dfrac{-q}{1-p-q}=\dfrac{q}{p+q-1}$

Hence,

$\implies\text{Product of zeroes =}\dfrac{q}{p+q-1} - - - -(6)$

We know that a cubic equation, can be expressed as: x³ - (Sum of zeroes)x² + (Sum of product of zeroes taken two at a time)x - (Product of zeroes) = 0.

So,

⇒ Cubic Equation : x³ - (Sum of zeroes)x² + (Sum of product of zeroes taken two at a time)x - (Product of zeroes) = 0

Using (4), (5) and (6),

$\implies\text{Cubic Equation =} x^3-\left(\dfrac{2p+3q}{p+q-1}\right)x^2+ \left(\dfrac{p+3q}{p+q-1}\right)x-\left(\dfrac{q}{p+q-1}\right)=0$

Taking LCM,

$\implies\text{Cubic Equation =} \dfrac{(p+q-1)x^3-(2p+3q)x^2+(p+3q)x-(q)}{p+q-1}=0$

Transposing (p + q -1) from Denominator to RHS,

$\implies\text{Cubic Equation =} (p+q-1)x^3-(2p+3q)x^2+(p+3q)x-q=0$

Therefore, the cubic equation whose roots are [α/(1+α)], [β/(1+β)] and [γ/(1+γ)] is (p + q - 1)x³ - (2p + 3q)x² + (p + 3q)x - q = 0.