Question

Find the zero of the quadratic polynomial $ {x^2} - 3 $ , and verify the relationship between the zeroes and coefficients.

Answer 1

Hint: For this type of problem we first find roots of a given quadratic equation and then we name them as $ \alpha \,\,and\,\,\beta $ . Either one will be named as alpha or beta and then equate the given quadratic equation with the standard equation to find the value of a, b and c and then use these values in relationship formulas to verify their relationship and hence the solution of the given problem.
Formulas used: Relationship between roots and coefficients: $ \alpha + \beta = - \dfrac{b}{a},\,\,\alpha .\beta = \dfrac{c}{a} $

Complete step-by-step answer:
Given quadratic polynomial $ {x^2} - 3 $
To find zeros of a given quadratic equation we equate to zero.
Therefore, we have
 $ {x^2} - 3 = 0 $
Or we can write above equation as:
 $ {x^2} - {\left( {\sqrt 3 } \right)^2} = 0 $
Now, writing in factor form by using an algebraic identity:
 $ {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right) $
Above equation can be written by using the above identity. We have,
\[
  \left( {x + \sqrt 3 } \right)\left( {x - \sqrt 3 } \right) = 0 \\
   \Rightarrow x + \sqrt 3 = 0\,\,\,or\,\,\,x - \sqrt 3 = 0 \\
   \Rightarrow x = - \sqrt 3 \,\,or\,\,x = \sqrt 3 \;
 \]
Therefore, from above we can say that the roots of the quadratic equation are $ \sqrt {3\,} \,\,and\, - \sqrt 3 $ .
So, the correct answer is “ $ \sqrt {3\,} \,\,and\, - \sqrt 3 $ .”.

Now, to verify the relationship between roots and coefficients.
Taking roots of quadratic equation as $ \alpha = \sqrt 3 \,\,and\,\,\beta = - \sqrt 3 $
Also, comparing the given quadratic equation $ {x^2} - 3 $ with the standard quadratic equation to get the value of a, b and c.
Hence, value of $ a = 1,\,\,b = 0\,\,and\,c = - 3 $
Now, writing the relationship between roots and coefficients.
We have
Sum of roots = $ - \dfrac{b}{a} $
Or we can write as:
 $ \alpha + \beta = - \dfrac{b}{a} $
Substituting values in above.
 $
  \sqrt 3 + \left( { - \sqrt 3 } \right) = - \dfrac{0}{1} \\
   \Rightarrow \sqrt 3 - \sqrt 3 = 0 \\
   \Rightarrow 0 = 0 \;
  $
Hence, sum of roots verified.
Now, writing product of roots = $ \dfrac{c}{a} $
Or we can write as:
 $ \alpha .\beta = \dfrac{c}{a} $
Now, substituting values in above. We have,
 $
  \sqrt 3 \left( { - \sqrt 3 } \right) = \dfrac{{ - 3}}{1} \\
   \Rightarrow - \sqrt 3 \times \sqrt 3 = - 3 \\
   \Rightarrow - 3 = - 3 \;
  $
Hence, the product of roots is also verified.
Therefore, from above we see that the relationship between roots and coefficients are verified.

Note: To factorise or to find roots of given quadratic equations there are different ways like middle term splitting method, quadratic formulas method but value of roots remains the same irrespective of formulas or method applied and roots are named as alpha and beta. Also, finding values of a, b and c from a given equation and then substituting values in their relationship between roots and coefficients to verify them and so solution of given problem.