Question

Find the zeroes of the following quadratic polynomial and verify the relationship between the zeroes and the coefficient. \[4{s^2} - 4s - 1 \]

Answer 1

Hint: We equate the given quadratic equation to zero. Then solving for \[x \] we have two roots for the given equation. If the given quadratic equation is in the form \[a{x^2} + bx + c = 0 \] then we know that the sum of zeros is the ratio of negative of coefficient of \[x \] to coefficient of \[{x^2} \] . We know that the product of zeros is the ratio of constant term to coefficient of \[{x^2} \] .

Complete step-by-step answer:
Given, \[4{s^2} - 4s - 1 \] , equating to zero.
 \[ \Rightarrow 4{s^2} - 4s - 1 = 0 \]
We cannot solve this by factorization method.
Now we know \[s = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \] , using this we can find roots.
Hear \[a = 4 \] , \[b = - 4 \] and \[c = - 1 \] . Substituting above we get,
 \[ \Rightarrow s = \dfrac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4(4)( - 1)} }}{{2(4)}} \]
Simplifying we get,
 \[ \Rightarrow s = \dfrac{{4 \pm \sqrt {16 + 16} }}{8} \]
 \[ \Rightarrow s = \dfrac{{4 \pm \sqrt {32} }}{8} \]
 \[ \Rightarrow s = \dfrac{{4 \pm 4 \sqrt 2 }}{8} \]
(Taking 4 as a common)
 \[ \Rightarrow s = \dfrac{{4(1 \pm \sqrt 2 )}}{8} \]
 \[ \Rightarrow s = \dfrac{{1 \pm \sqrt 2 }}{2} \]
Thus, the two roots are the given quadratic equation is
 \[ \Rightarrow s = \dfrac{{1 + \sqrt 2 }}{2} \] and \[ \Rightarrow s = \dfrac{{1 - \sqrt 2 }}{2} \] .
Now from standard definition and derivation we have,
Sum of zeros \[ = - \dfrac{b}{a} = - \dfrac{{ - 4}}{4} = 1 \]
Product of zeros \[ = \dfrac{c}{a} = \dfrac{{ - 1}}{4} = - \dfrac{1}{4} \] .
Now, taking the roots of the above quadratic equation and solving.
Sum of zeros \[ \Rightarrow s = \dfrac{{1 + \sqrt 2 }}{2} + \dfrac{{1 - \sqrt 2 }}{2} \]
 \[ \Rightarrow s = \dfrac{{1 + \sqrt 2 + 1 - \sqrt 2 }}{2} = 1 \]
Product of zeros \[ \Rightarrow s = \dfrac{{1 + \sqrt 2 }}{2} \times \dfrac{{1 - \sqrt 2 }}{2} \]
 \[ \Rightarrow s = \dfrac{{1 - \sqrt 2 + \sqrt 2 - 2}}{4} = - \dfrac{1}{4} \]
We can see in both cases that finding the value of the sum of zeros and product of zeros are the same.
Hence it satisfies the relation between zeros and coefficients of a quadratic equation.
So, the correct answer is “\[ \Rightarrow s = \dfrac{{1 + \sqrt 2 }}{2} \] and \[ \Rightarrow s = \dfrac{{1 - \sqrt 2 }}{2} \] .”.

Note: Don’t get confused with zeros of quadratic equations and roots of quadratic equations. Both are the same. Above all we do is simple multiplication and addition. Remember this method is the same for any quadratic equations. Substitute the given any equation to zero so that we are able to solve it.