Question

How do you find a quadratic polynomial with integer coefficient which has $x = \dfrac{3}{5} \pm \dfrac{{\sqrt {29} }}{5}$ as its real zeros ?

Answer 1

Hint: In this question we are given the zeros (roots) of the quadratic polynomial and from the given zeros we need to find the quadratic polynomial with integer coefficient.
Make use of the relation between the zeros and the coefficient of a polynomial to find the required quadratic equation. We make use of the formula ${x^2} - (\alpha + \beta )x + \alpha \beta = 0$ where $\alpha $ and $\beta $ are zeros of the quadratic equation. Also try to draw some results seeing the nature of the roots.

Complete step by step solution:
Given the roots (zeros) of the quadratic equation, $x = \dfrac{3}{5} \pm \dfrac{{\sqrt {29} }}{5}$
We know that for a general quadratic equation $a{x^2} + bx + c = 0$ the roots are given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
 Let $\alpha $ and $\beta $ be roots of the quadratic equation.
The relation between the coefficient and the roots of a general quadratic equation comes out to be,
Sum of the roots of the quadratic equation, $\alpha + \beta = \dfrac{{ - b}}{a}$
Product of the roots of the equation, $\alpha \beta = \dfrac{c}{a}$
Hence the quadratic equation with its roots can be written as,
$k({x^2} - (\alpha + \beta )x + \alpha \beta ) = 0$ ……(1)
Where k is a constant and can have value according to our need.
In the above question we have,
$\alpha = \dfrac{3}{5} + \dfrac{{\sqrt {29} }}{5}$
$\beta = \dfrac{3}{5} - \dfrac{{\sqrt {29} }}{5}$
Now we calculate $\alpha + \beta $ and $\alpha \beta $, then substitute these values in the equation (1) to get the desired quadratic equation.
$\alpha + \beta = \dfrac{3}{5} + \dfrac{{\sqrt {29} }}{5} + \dfrac{3}{5} - \dfrac{{\sqrt {29} }}{5}$
Computing like terms $\dfrac{{\sqrt {29} }}{5} - \dfrac{{\sqrt {29} }}{5} = 0$.
$ \Rightarrow \alpha + \beta = 0 + \dfrac{3}{5} + \dfrac{3}{5}$
$ \Rightarrow \alpha + \beta = \dfrac{3}{5} + \dfrac{3}{5}$
$ \Rightarrow \alpha + \beta = \dfrac{6}{5}$
Now $\alpha \beta = \left( {\dfrac{3}{5} + \dfrac{{\sqrt {29} }}{5}} \right)\left( {\dfrac{3}{5} - \dfrac{{\sqrt {29} }}{5}} \right)$
This is of the form $(a + b)(a - b)$.
We have the formula, $({a^2} - {b^2}) = (a + b)(a - b)$
Here $a = \dfrac{3}{5}$ and $b = \dfrac{{\sqrt {29} }}{5}$.
$ \Rightarrow \alpha \beta = \left( {{{\left( {\dfrac{3}{5}} \right)}^2} - {{\left( {\dfrac{{\sqrt {29} }}{5}} \right)}^2}} \right)$
$ \Rightarrow \alpha \beta = \left( {\left( {\dfrac{9}{{25}}} \right) - \left( {\dfrac{{29}}{{25}}} \right)} \right)$
$ \Rightarrow \alpha \beta = \dfrac{{ - 20}}{{25}}$
$ \Rightarrow \alpha \beta = - \dfrac{4}{5}$
Now substituting the values of $\alpha + \beta $and $\alpha \beta $ in the equation (1), we get,
$k({x^2} - (\alpha + \beta )x + \alpha \beta ) = 0$
$ \Rightarrow k\left( {{x^2} - \dfrac{6}{5}x + \left( { - \dfrac{4}{5}} \right)} \right) = 0$
Now, as no condition given in the question, we let k to be equal to 1.
Hence the above equation becomes,
$\Rightarrow 1 \cdot \left( {{x^2} - \dfrac{6}{5}x + \left( { - \dfrac{4}{5}} \right)} \right) = 0$
$ \Rightarrow \left( {{x^2} - \dfrac{6}{5}x - \dfrac{4}{5}} \right) = 0$
Now taking LCM in the L.H.S. we get,
$ \Rightarrow \dfrac{{5{x^2} - 6x - 4}}{5} = 0$
Now taking 5 to the other side we get,
$ \Rightarrow 5{x^2} - 6x - 4 = 0 \times 5$
$ \Rightarrow 5{x^2} - 6x - 4 = 0$

Therefore, we can conclude that for a given zeros $x = \dfrac{3}{5} \pm \dfrac{{\sqrt {29} }}{5}$ one of the required quadratic equation with integer coefficient is $5{x^2} - 6x - 4 = 0$.

Note:
We must remember that we can form an infinite number of quadratic equations when both of its roots are given. So to get more quadratic equations we need to do is keep on changing the value of k in the equation given by,
$k({x^2} - (\alpha + \beta )x + \alpha \beta ) = 0$, where $\alpha $ and $\beta $ are zeros of the quadratic equation.
Also remember that for a quadratic equation with integer coefficients as in the above problem, irrational or complex roots always appear in conjugate pairs.