Question

find the quadractic polynomial whose zeros are 3/4 and 1/4 ​

Answer 1

$\large{\underline{\green{\bf{Solution:}}}}$

We are given two zeroes of an unknown polynomial. Let us assume:

• $\sf{\alpha = \dfrac{3}{4}}$

• $\sf{\beta = \dfrac{1}{4}}$

Let us find the sum of zeroes of the polynomial.

$\longmapsto\sf{\alpha + \beta = \dfrac{3}{4}+ \dfrac{1}{4}}$

$\longmapsto\sf{\alpha + \beta = \dfrac{4}{4}}$

$\longmapsto\boxed{\bf{\alpha + \beta = 1}}$

Also, it is necessary to find the product of zeroes:

$\longmapsto\sf{\alpha\beta = \dfrac{3}{4} \times \dfrac{1}{4}}$

$\longmapsto\boxed{\bf{\alpha\beta = \dfrac{3}{16}}}$

To find the quadratic polynomial, we use the formula:

$\longrightarrow\tt{k[x^2-(Sum)x+Product]}$

$\longrightarrow\tt{k[x^2-(\alpha + \beta)x+ \alpha\beta]}$

By applying our values, we get:

$\longrightarrow\tt{k[x^2-(1)x+ \dfrac{3}{16}]}$

$\longrightarrow\tt{k[x^2-x+ \dfrac{3}{16}]}$

Let us remove the denominator in the constant by assuming k as 16. We can also say that, we are multiplying 16 to the whole polynomial.

$\longrightarrow\tt{16[x^2-x+ \dfrac{3}{16}]}$

$\longrightarrow\boxed{\bf{16x^2-16x+3}}$

Hence, this is our required polynomial.

$\large{\underline{\green{\bf{Note:}}}}$

You can also find the polynomial by the following way in which:

• $\sf{\alpha + \beta = \dfrac{-b}{a}}$

• $\sf{\alpha\beta = \dfrac{c}{a}}$

From this relation too, we can find the unknown polynomial.