Question

Find the zeros of the quadratic polynomial ${x^2} - 2x - 8$ and verify the relationship between the zeros and the coefficients.

Answer 1

Hint: The given equation is a quadratic equation and for finding the zeros of the quadratic equation, we will be using the quadratic formula.
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
After finding the zeros, we need to verify the two relations that we have between the zeros of the quadratic equation and the coefficients of the equation.
Sum of zeros $\left( {\alpha + \beta } \right) = \dfrac{{ - b}}{a}$
Product of zeros $\left( {\alpha \beta } \right) = \dfrac{c}{a}$

Complete step-by-step answer:
In this question, we are given a quadratic polynomial and we are supposed to find its zeros and then verify the relation between its zeros and the coefficients of the quadratic equation.
Given equation: ${x^2} - 2x - 8$
Zero of a equation means a value of the variable that when substituted in the equation, gives the value of the equation as 0.
Now, to find the zeros of a given equation, we have to equate the equation with 0.
Therefore, equating our given equation with 0, we get
${x^2} - 2x - 8 = 0$
Now, using the quadratic formula that is
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Where, $a = 1,b = - 2,c = - 8$
Therefore,
\[
   \Rightarrow x = \dfrac{{ - \left( { - 2} \right) \pm \sqrt {{{\left( { - 2} \right)}^2} - 4\left( 1 \right)\left( { - 8} \right)} }}{{2\left( 1 \right)}} \\
   \Rightarrow x = \dfrac{{2 \pm \sqrt {4 - \left( { - 32} \right)} }}{2} \\
   \Rightarrow x = \dfrac{{2 \pm \sqrt {4 + 32} }}{2} \\
   \Rightarrow x = \dfrac{{2 \pm \sqrt {36} }}{2} \\
   \Rightarrow x = \dfrac{{2 \pm 6}}{2} \\
 \]
$ \Rightarrow x = \dfrac{{2 + 6}}{2} = \dfrac{8}{2} = 4$ and $ \Rightarrow x = \dfrac{{2 - 6}}{2} = \dfrac{{ - 4}}{2} = - 2$
Hence, the two zeros of the quadratic equation ${x^2} - 2x - 8$ are 4 and -2.
Now, there are two relations between the zeros of the equation and the coefficients of the equation.
Sum of zeros $\left( {\alpha + \beta } \right) = \dfrac{{ - b}}{a}$
Product of zeros $\left( {\alpha \beta } \right) = \dfrac{c}{a}$
So, in our equation,
$\alpha = 4$
$\beta = - 2$
$a = 1$
$b = - 2$
$c = - 8$
Let us verify both the equations by putting these values in them.
$ \Rightarrow $Sum of zeros:
$\alpha + \beta = \dfrac{{ - b}}{a}$
$
  4 + \left( { - 2} \right) = \dfrac{{ - \left( { - 2} \right)}}{1} \\
  4 - 2 = \dfrac{2}{1} \\
  2 = 2 \\
 $
Hence, LHS=RHS.
Therefore, the first relation is verified.
$ \Rightarrow $Product of zeros:
$\left( {\alpha \beta } \right) = \dfrac{c}{a}$
$
  \left( 4 \right)\left( { - 2} \right) = \dfrac{{ - 8}}{1} \\
   - 8 = - 8 \\
 $
Hence, LHS=RHS.
Therefore, the second relationship is also verified.

Note: Here, we can check that whether our zeros of the equation are correct or not by putting those values in the give equation and if the answer will be 0, then our zeros are correct and if not, then they must be wrong.
$
   \Rightarrow {x^2} - 2x - 8 \\
   \Rightarrow {\left( 4 \right)^2} - 2\left( 4 \right) - 8 \\
   \Rightarrow 16 - 16 \\
   \Rightarrow 0 \\
 $ $
   \Rightarrow {\left( { - 2} \right)^2} - 2\left( { - 2} \right) - 8 \\
   \Rightarrow 4 + 4 - 8 \\
   \Rightarrow 8 - 8 \\
   \Rightarrow 0 \\
 $
Hence, both the values of the zeros are correct.