Question

If a and b are the roots of the equation \[{{x}^{2}}-1=0\] the form of quadratic equation whose roots are 2a and 2b will be:
A. \[{{x}^{2}}-4\sqrt{2}=0\]
B. \[{{x}^{2}}+4=0\]
C. \[{{x}^{2}}-4=0\]
D. \[{{x}^{2}}+4\sqrt{2}=0\]

Answer 1

Hint: First of all we will find the roots of the quadratic equation and thus we will get the value of ‘a’ and ‘b’ and then we will check the options one by one whose roots will be 2a and 2b.

Complete step-by-step answer:

We have been given a and b are the roots of the equation \[{{x}^{2}}-1=0\] and asked to form the quadratic equations whose roots are 2a and 2b.
\[\Rightarrow {{x}^{2}}-1=0\]
Adding 1 to both sides of equation, we get as follows:
\[\begin{align}
  & \Rightarrow {{x}^{2}}-1+1=1 \\
 & \Rightarrow {{x}^{2}}=1 \\
\end{align}\]
Taking square roots on both the sides, we get as follows:
\[\begin{align}
  & \Rightarrow x=\sqrt{1} \\
 & \Rightarrow x=1,x=-1 \\
\end{align}\]
Hence a=1 and b=-1.
So, 2a=2 and 2b=-2.
Now we will check the options one by one as follows to check the quadratic equation whose roots are 2a and 2b.

A. \[{{x}^{2}}-4\sqrt{2}=0\]
Adding \[4\sqrt{2}\] to both the sides of the equation, we get as follows:
\[\begin{align}
  & {{x}^{2}}-4\sqrt{2}+4\sqrt{2}=4\sqrt{2} \\
 & \Rightarrow {{x}^{2}}=4\sqrt{2} \\
\end{align}\]
Taking square roots on both the sides, we get as follows:
\[\Rightarrow x=\sqrt{4\sqrt{2}}=\pm 2\sqrt{\sqrt{2}}\]
Hence this option is not correct.

B. \[{{x}^{2}}+4=0\]
Subtracting 4 from both the sides, we get as follows:
\[\begin{align}
  & {{x}^{2}}+4-4=-4 \\
 & \Rightarrow {{x}^{2}}=-4 \\
\end{align}\]
Taking square root on both the sides, we get as follows:
\[\begin{align}
  & x=\pm \sqrt{-4} \\
 & x=2i,x=-2i \\
\end{align}\]
Where ‘i’ is known as iota and equal to \[\sqrt{-1}\].
Hence this option is not correct.

C. \[{{x}^{2}}-4=0\]
Adding 4 to both the sides, we get as follows:
\[\begin{align}
  & {{x}^{2}}-4+4=4 \\
 & \Rightarrow {{x}^{2}}=4 \\
\end{align}\]
Taking square root on both the sides, we get as follows:
\[\begin{align}
  & x=\pm \sqrt{4} \\
 & x=2,x=-2 \\
\end{align}\]
Hence this option is correct.

D. \[{{x}^{2}}+4\sqrt{2}=0\]
Subtracting \[4\sqrt{2}\] from both the sides of the equation, we get as follows:
\[\begin{align}
  & {{x}^{2}}+4\sqrt{2}-4\sqrt{2}=-4\sqrt{2} \\
 & \Rightarrow {{x}^{2}}=-4\sqrt{2} \\
\end{align}\]
Taking square root on both the sides, we get as follows:
\[x=\pm \sqrt{-4\sqrt{2}}\]
Hence this option is not correct.
Therefore, the only correct option of the given question is option C.

Note: We can solve this question by another method, since 2a and 2b are the roots of a quadratic equation then according to the factor theorem we can say that (x-2a) and (x-2b) are the factors of the quadratic equation. Thus we will get the equation by multiplying its factors i.e. \[\left( x-2a \right)\left( x-2b \right)=0\].