Question

Find the zeros of the polynomial $f(x)={{x}^{2}}-2$ and verify the relationship between its zeros and coefficients.

Answer 1

Hint: First compare it to conditions of equations,\[a{{x}^{2}}+bx+c\]. Now apply all the conditions which are applicable to the general equation verify the conditions of the sum of zeroes, product of zeroes. This verification is your required result to this question.

Complete step-by-step answer:
Given function in the question, can be written in form:
$f(x)={{x}^{2}}-2$
To find zeros of function, we must equate it to 0:
${{x}^{2}}-2=0$………… (i)
General quadratic equations will be in the form of:
$a{{x}^{2}}+bx+c=0$………… (ii)
 The zeroes are nothing but the roots of equation given:
The zeroes of equation (ii), can be written by formula as:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$……….. (iii)
The sum of zeros of equation (ii), satisfy relation given by:
Sum of zeroes =$\dfrac{-b}{a}$ ……… (iv)
The product of zeroes of equation (ii) satisfy relation given by
Product of zeroes =$\dfrac{c}{a}$……… (v)
By comparing the equations (i), (ii), we can say the following:
The value of variable a is found out to be 1.
The value of variable b is found out to be 0.
The value of variable c is found out to be -2.
 By substituting these values in the equation (iii), we get it as:
$x=\dfrac{-0\pm \sqrt{{{0}^{2}}-4(1)(-2)}}{2(1)}$
By Simplifying the equation, we can write it in form of
$x=\pm \dfrac{\sqrt{4(2)}}{2}$
By bringing 4 outside the square root, we can write it as:
$x=\pm \dfrac{2\sqrt{2}}{2}$
By simplifying and writing them separately, we get
Zeroes as $\sqrt{2}$,$-\sqrt{2}$
By substituting all value in the equation (iv), we get it as:
$\sqrt{2}+\left( -\sqrt{2} \right)=\dfrac{0}{1}\Rightarrow 0=0$
As LHS=RHS, we verified this equation from $\sqrt{2}$,$-\sqrt{2}$.
By substituting all values in the equation (v), we get it as:
$(\sqrt{2})(-\sqrt{2})=\dfrac{-2}{1}\Rightarrow -2=-2$
As LHS =RHS, we also verified the product of zeroes, by this.
Therefore the roots are$\sqrt{2}$,$-\sqrt{2}$ and we verified conditions.

Note: You can directly add 2 on both sides and then apply square root on both sides to get a result, but verification of relations of roots, coefficient should be done only in this process. So, it will look better if you find zeroes also using this method. Be careful while verifying the relations as it is the base for solution. Take care of +,- sign while substituting them.