Write the quadratic equation whose zeroes are \[2\]and \[ - 4\].

Hint: In the given question we have to find the quadratic equation where we have been given two zeroes \[2\]and \[ - 4\]. Now first of all you should know the standard form of a quadratic equation which is \[a{x^2} + bx + c\].
Now when write a quadratic equation using roots we get:
\[k[{x^2} - (\alpha + \beta )x + (\alpha \times \beta )]\]
Where, \[\alpha ,\beta \]are the two zeroes or roots of the equation and k is any constant.
Now you can easily find the quadratic equation whose zeroes are \[2\]and \[ - 4\].

Complete step-by-step answer:
In the given question we have to find the quadratic equation where we have been given two zeroes \[2\]and \[ - 4\].
Now we have a general form of a quadratic equation which is formed using the two zeroes or two roos of the quadratic equation i.e.
\[k[{x^2} - (\alpha + \beta )x + (\alpha \times \beta )]\]
Here we have, \[\alpha = 2,\beta = - 4\]
There putting these values in the general form we get:
\[
  k[{x^2} - (2 - 4)x + (2 \times - 4)] \\
   = k[{x^2} - ( - 2)x + ( - 8)] \\
   = k[{x^2} + 2x - 8] \\
 \]
Now to find a quadratic equation take \[k = 1\]we get:
\[
  1 \times [{x^2} + 2x - 8] \\
   = {x^2} + 2x - 8 \\
 \]

Note: Here you should remember the general form of the quadratic equation and also the general form where we have used the roots or zeros of the quadratic equation. Also you should remember that the zeroes is also known as the root of the quadratic equation.
Here I am mentioning the general form of a quadratic equation and how to find the roots of a quadratic equation:
\[a{x^2} + bx + c\]
Now let roots of the above given quadratic equation is \[\alpha ,\beta \], so we can also write\[\alpha ,\beta \]as:
\[\alpha = \dfrac{{ - b}}{a},\beta = \dfrac{c}{a}\]
Also you can form a quadratic equation using these roots i.e.
\[k[{x^2} - (\alpha + \beta )x + (\alpha \times \beta )]\]