Question

Find a quadratic polynomial whose zeroes are $\dfrac{{3 - \sqrt 3 }}{5}$ and $\dfrac{{3 + \sqrt 3 }}{5}$.

Answer 1

Hint: In this question, we are given two zeroes of a quadratic and we have been asked to find that quadratic polynomial. Using the two zeroes, find the sum of the zeroes and then find the product. Put both of them in the equation ${x^2} - \left( {\alpha + \beta } \right)x + \alpha \beta $. You will get an equation. Simplify the equation by taking a common denominator. The resultant equation will be a quadratic polynomial.

Formula used: ${x^2} - \left( {\alpha + \beta } \right)x + \alpha \beta $

Complete step-by-step solution:
Let us first know a little about quadratic equations.
Quadratic equation is an equation in one variable, having degree 2. One thing to be understood is that the number of solutions of an equation is equal to the degree of the equation.
Some important formula to remember about the quadratic equation ${x^2} + bx + c = 0$ are –
1) $\alpha + \beta = \dfrac{{ - b}}{a}$
2) $\alpha \beta = \dfrac{c}{a}$
We are given the two zeroes of the quadratic polynomial - $\dfrac{{3 - \sqrt 3 }}{5}$ and $\dfrac{{3 + \sqrt 3 }}{5}$. Let us assume $\alpha = \dfrac{{3 - \sqrt 3 }}{5}$ and $\beta = \dfrac{{3 + \sqrt 3 }}{5}$.
Now, we will find the sum and product of the two zeroes –
Sum of zeroes = $\alpha + \beta $
$ \Rightarrow \alpha + \beta = \dfrac{{3 - \sqrt 3 }}{5} + \dfrac{{3 + \sqrt 3 }}{5}$
On simplifying we will get,
$ \Rightarrow \alpha + \beta = \dfrac{{3 - \sqrt 3 + 3 + \sqrt 3 }}{5}$
$ \Rightarrow \alpha + \beta = \dfrac{6}{5}$
Product of zeroes = $\alpha \beta $
$ \Rightarrow \alpha \beta = \left( {\dfrac{{3 - \sqrt 3 }}{5}} \right)\left( {\dfrac{{3 + \sqrt 3 }}{5}} \right)$
$ \Rightarrow \alpha \beta = \dfrac{{\left( {3 - \sqrt 3 } \right)\left( {3 + \sqrt 3 } \right)}}{{5 \times 5}}$
On simplifying we will get,
$ \Rightarrow \alpha \beta = \dfrac{{9 + 3\sqrt 3 - 3\sqrt 3 - 3}}{{25}}$
$ \Rightarrow \alpha \beta = \dfrac{6}{{25}}$
Now, we will put the sum and the product of the zeroes in the equation.
$ \Rightarrow {x^2} - \left( {\alpha + \beta } \right)x + \alpha \beta $
Putting all the values,
$ \Rightarrow {x^2} - \left( {\dfrac{6}{5}} \right)x + \dfrac{6}{{25}}$$ = 0$
Simplifying the equation,
$ \Rightarrow \dfrac{{{x^2} \times 25}}{{25}} - \left( {\dfrac{{6 \times 5}}{{5 \times 5}}} \right)x + \dfrac{6}{{25}}$$ = 0$
$ \Rightarrow 25{x^2} - 30x + 6 = 0$

Therefore, the quadratic polynomial from the given two zeroes is $25{x^2} - 30x + 6 = 0$.

Note: The step where we find the product of the zeroes, instead of finding the product by distributive property, we can also use the formula of $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$ like this –
$ \Rightarrow \alpha \beta = \left( {\dfrac{{3 - \sqrt 3 }}{5}} \right)\left( {\dfrac{{3 + \sqrt 3 }}{5}} \right)$
$ \Rightarrow \alpha \beta = \dfrac{{\left( {3 - \sqrt 3 } \right)\left( {3 + \sqrt 3 } \right)}}{{5 \times 5}}$
Here, we can put $a = 3,b = \sqrt 3 $ and use the formula $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$.
$ \Rightarrow \left( {3 + \sqrt 3 } \right)\left( {3 - \sqrt 3 } \right) = {3^2} - {\left( {\sqrt 3 } \right)^2}$
$ \Rightarrow 9 - 3 = 6$
$ \Rightarrow \alpha \beta = \dfrac{6}{{25}}$