Question

Find a quadratic polynomial with \[4\] and \[5\] as the zeroes of the polynomial
(A) $ {x^2} - 9x + 20 $
(B) $ {x^2} - 5x + 9 $
(C) $ {x^2} - 4x + 5 $
(D) None of these.

Answer 1

Hint: Before proceeding further we need to know that if $ \alpha \,{\text{and}}\,\,\beta $ are the roots of the equation then the quadratic equation so formed will be represented by-
\[
  {x^2} - \left( {{\text{Sum of Roots}}} \right)x + \left( {{\text{Product of Roots}}} \right) \\
\Rightarrow \,{x^2} - \left( {\alpha {\text{ + }}\beta } \right)x + \left( {\alpha \beta } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right)
\]

Complete step-by-step answer:
A root or a zero of a polynomial are the value(s) of X that cause the polynomial to = 0 (or make Y=0). It is an X-intercept. The root is the X-value, and zero is the Y-value.
Remember that a quadratic polynomial (we assume that the coefficients are real) will always have two zeroes, but the nature of the zeroes depends on the coefficients:
Now according to this question we have,
$
  \alpha = 4 \\
  \beta = 5
$
Substitute the values in equation\[\left( 1 \right)\], we have,
\[
\Rightarrow \,{x^2} - \left( {4 + 5} \right)x + \left( {4 \times 5} \right) \\
\Rightarrow {x^2} - 9x + 20
\]
 So by the discussion, we see that the quadratic polynomial is \[{x^2} - 9x + 20\].

So, the correct answer is “Option A”.

Note: While writing the roots of the equation we must not change the sign. Write the sign as it is mentioned in the question. If there are two zeroes or there are two roots of the equation, then a quadratic equation is formed.