Question

Find the zero of the polynomial \[{{x}^{2}}-3x\].

Answer 1

Hint: Now we have to compare \[a{{x}^{2}}+bx+c=0\] with \[{{x}^{2}}-3x=0\]. We have to find the values of a, b and c. Let us assume the roots of equation \[a{{x}^{2}}+bx+c=0\] are \[\alpha \] and \[\beta \]. We know the roots of a quadratic equation \[a{{x}^{2}}+bx+c=0\] are equal to \[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]. Let us assume\[\alpha =\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a},\beta =\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}\]. Now we have to put the values of a, b and c in \[\alpha =\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a},\beta =\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}\]. The values of \[\alpha \]and \[\beta \] give us the value of roots of\[a{{x}^{2}}+bx+c=0\].

Complete step-by-step answer:
Let us compare the quadratic equation \[a{{x}^{2}}+bx+c=0\] with \[{{x}^{2}}-3x=0\].
\[\begin{align}
  & a=1......(1) \\
 & b=-3....(2) \\
 & c=0.....(3) \\
\end{align}\]
Let us assume the roots of equation \[a{{x}^{2}}+bx+c=0\] are \[\alpha \] and \[\beta \].
We know the roots of a quadratic equation \[a{{x}^{2}}+bx+c=0\] are equal to \[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
Let us assume
\[\alpha =\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a},\beta =\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}\]
Now let us assume the roots of a quadratic equation \[{{x}^{2}}-3x=0\] are \[\alpha \] and \[\beta \].
From equation (1), equation (2) and equation (3), we get
\[\begin{align}
  & \Rightarrow \alpha =\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a} \\
 & \Rightarrow \alpha =\dfrac{-(-3)+\sqrt{{{(-3)}^{2}}-4(1)(0)}}{2(1)} \\
 & \Rightarrow \alpha =\dfrac{-(-3)+\sqrt{9-0}}{2(1)} \\
 & \Rightarrow \alpha =\dfrac{-(-3)+\sqrt{9}}{2(1)} \\
 & \Rightarrow \alpha =\dfrac{-(-3)+3}{2(1)} \\
 & \Rightarrow \alpha =\dfrac{3+3}{2(1)} \\
 & \Rightarrow \alpha =\dfrac{3+3}{2} \\
 & \Rightarrow \alpha =\dfrac{6}{2} \\
 & \Rightarrow \alpha =3.....(4) \\
\end{align}\]
\[\begin{align}
  & \Rightarrow \beta =\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a} \\
 & \Rightarrow \beta =\dfrac{-(-3)-\sqrt{{{(-3)}^{2}}-4(1)(0)}}{2(1)} \\
 & \Rightarrow \beta =\dfrac{-(-3)-\sqrt{9-0}}{2(1)} \\
 & \Rightarrow \beta =\dfrac{-(-3)-\sqrt{9}}{2(1)} \\
 & \Rightarrow \beta =\dfrac{-(-3)-3}{2(1)} \\
 & \Rightarrow \beta =\dfrac{3-3}{2(1)} \\
 & \Rightarrow \beta =\dfrac{3-3}{2} \\
 & \Rightarrow \beta =\dfrac{0}{2} \\
 & \Rightarrow \beta =0.....(5) \\
\end{align}\]
From equation (4) and equation (5), we get
The roots of \[{{x}^{2}}-3x=0\] are 0 and 3.

Note: This problem can be solved in an alternative manner.
Let us assume the roots of equation \[a{{x}^{2}}+bx+c=0\] are \[\alpha \] and \[\beta \], then we get
\[\begin{align}
  & \alpha +\beta =\dfrac{-b}{a}......(1) \\
 & \alpha \beta =\dfrac{c}{a}......(2) \\
\end{align}\]
Let us compare the quadratic equation \[a{{x}^{2}}+bx+c=0\] with \[{{x}^{2}}-3x=0\].
\[\begin{align}
  & a=1......(3) \\
 & b=-3....(4) \\
 & c=0.....(5) \\
\end{align}\]
Now let us assume the roots of a quadratic equation \[{{x}^{2}}-3x=0\] are \[\alpha \] and \[\beta \].
Now we will substitute equation (3) and equation (4) in equation (5). Then we will get
\[\begin{align}
  & \alpha +\beta =\dfrac{-(-3)}{1} \\
 & \Rightarrow \alpha +\beta =3......(6) \\
 & \alpha \beta =\dfrac{0}{1} \\
 & \Rightarrow \alpha \beta =0......(7) \\
\end{align}\]
Let us assume \[\alpha =0\].
Now we will substitute \[\alpha =0\]in equation (6), we get
\[\begin{align}
  & \Rightarrow 0+\beta =3 \\
 & \Rightarrow \beta =3.......(8) \\
\end{align}\]
From equation (7) and equation (8), we get 0 and 3 are the roots of \[{{x}^{2}}-3x=0\].