Question

Find a quadratic polynomial whose sum and product respectively of the zeroes are given. Also, find the zeroes of this polynomial by factorization.
$ - 2\sqrt 3 , - 9$

Answer 1

Hint: Here we go through by assuming the general quadratic equation $a{x^2} + bx + c$ and let their zeroes as \[\alpha \] and $\beta $. Then apply the properties of quadratic equation i.e. $\alpha + \beta = - \dfrac{b}{a}$ and $\alpha \beta = \dfrac{c}{a}$. By using these properties, we will get the equation. After that use the factorization method to get the zeroes.

Complete step-by-step solution:
Let the quadratic polynomial be $a{x^2} + bx + c$ and their zeroes as \[\alpha \] and $\beta $.
And we know that by the properties of the quadratic equation the sum of zeroes is equal to the negative of the coefficient of x divided by the coefficient of ${x^2}$ i.e., $\alpha + \beta = - \dfrac{b}{a}$.
And the product of the zeroes is equal to the constant term divided by the coefficient of ${x^2}$ i.e., $\alpha \beta = \dfrac{c}{a}$.
Now in the question, it is given that the sum of the zeroes is $ - 2\sqrt 3 $ that we can write as
$ \Rightarrow - \dfrac{b}{a} = - 2\sqrt 3 $..................….. (1)
And the product of the zeroes is given as $ - 9$ that we can write as,
$ \Rightarrow \dfrac{c}{a} = - 9$...............….. (2)
And we know that the general quadratic equation can be written as
${x^2} - $(sum of zeroes)$x + $(product of zeroes)
By equation (1) and equation (2), we can identify the sum of zeroes and the product of zeroes.
Now put these values in the general form of a quadratic equation we get it as,
$ \Rightarrow {x^2} - \left( { - 2\sqrt 3 } \right)x + \left( { - 9} \right) = 0$
Simplify the terms,
$\therefore {x^2} + 2\sqrt 3 x - 9 = 0$
So, the quadratic polynomial is ${x^2} + 2\sqrt 3 x - 9 = 0$.
Now factorize it by factorization method,
We can write $2\sqrt 3 $ as $3\sqrt 3 - \sqrt 3 $,
$ \Rightarrow {x^2} + \left( {3\sqrt 3 - \sqrt 3 } \right)x - 9 = 0$
Open the brackets and multiply the terms,
$ \Rightarrow {x^2} + 3\sqrt 3 x - \sqrt 3 x - 9 = 0$
Take common factors from the equation,
$ \Rightarrow x\left( {x + 3\sqrt 3 } \right) - \sqrt 3 \left( {x + 3\sqrt 3 } \right) = 0$
Take common factors from the equation,
$ \Rightarrow \left( {x + 3\sqrt 3 } \right)\left( {x - \sqrt 3 } \right) = 0$
Then,
$\therefore x = - 3\sqrt 3 ,\sqrt 3 $

Hence, the quadratic equation is ${x^2} + 2\sqrt 3 x - 9 = 0$ and the zeroes are $ - 3\sqrt 3 ,\sqrt 3 $.

Note: Whenever we face such a type of question the key concept for solving the question is. Always assume the quadratic equation in its general form and let its zeroes as $\alpha $ and $\beta $ then go according to the question and equate it to get the values of sum of roots and product of roots and then put the values of these in general form to get the required quadratic equation. And always take care of signs. We generally make mistakes here.