Question

Find the zeros of the quadratic polynomial, $p(x) = 4{x^2} + 24x + 36$ and verify the relationship between the zeros and their coefficient.

Answer 1

Hint: A quadratic equation is represented by $p(x) = a{x^2} + bx + c$, where a, b, c are the coefficients. To find the zeros of the quadratic polynomial, equate $p(x) = 0$.

Complete step-by-step answer:
The given equation is, $p(x) = 4{x^2} + 24x + 36$.

To find the zeros, put $p(x) = 0$

$ \Rightarrow 4{x^2} + 24x + 36 = 0$

Dividing the equation by 4, we get

$ \Rightarrow {x^2} + 6x + 9 = 0$

Solving the equation further,

$

   \Rightarrow {x^2} + 3x + 3x + 9 = 0 \\

   \Rightarrow x(x + 3) + 3(x + 3) = 0 \\

   \Rightarrow (x + 3)(x + 3) = 0 \\

   \Rightarrow x = - 3, - 3 \\

$

Hence, the roots of the given equation are, $\alpha = - 3,\beta = - 3$.

Now, if we find the sum of roots i.e., $\alpha + \beta = - 3 + ( - 3) = - 6 - (1)$

And the product of the roots i.e., $\alpha .\beta = - 3( - 3) = 9 - (2)$

Now the coefficients of the given quadratic eq are-

$a = 4,b = 24,c = 36$

If we find, $ - \dfrac{b}{a} = - \dfrac{{24}}{4} = 6 = \alpha + \beta $ and $\dfrac{c}{a} =

\dfrac{{36}}{4} = 9 = \alpha .\beta $.

Hence, the zeros of the polynomial are $\alpha = - 3,\beta = - 3$ and the relationship

between the zeros and their coefficients is-

\[

  \alpha + \beta = \dfrac{{ - b}}{a} \\

  \alpha .\beta = \dfrac{c}{a} \\

\]

Note: In such types of questions, always find the zeros of the polynomial first, if it is a quadratic polynomial then the no. of zeros will be two. Then, find the sum of the zeros and the product of the roots. Find the ratios of the coefficients and then see their relationship with the zeros of the polynomial.