Question

How do you solve \[16{{x}^{2}}-81\]?

Answer 1

Hint: We are given \[16{{x}^{2}}-81\], we are asked to find the factor of this, to do so we will first understand the type of equation we have, once we get that we reduce the given equation into the standard form to simplify after that we will find the greatest common factor from each term then in the remaining term be factor using the middle term and lastly, we compare it with zero and solve further.

Complete step by step solution:
We use \[a\times c\] in such a way that its sum or difference from the ‘b’ of the equation \[a{{x}^{2}}+bx+c=0\] We are given \[16{{x}^{2}}-81\] we are asked to find the factor of it. Now to find the factor of the equation, we should see that as the highest power is \[2\]so it is or \[2\]-degree polynomial. So, it is a quadratic equation. As our Equation is quadratic equations and we know quadratic equations are given as \[a{{x}^{2}}+bx+c=0\]. We reduce it to the standard form
We can see that our equation \[16{{x}^{2}}-81\] is already in standard form. Now, to find its factor we will first find the possible greatest common factor of all these. In \[16\text{ and }81\] we can see that \[1\] is the only possible term that can be separated, so we get – our equation stays like original as \[-16{{x}^{2}}-81\]
We can write this above equation as
\[16{{x}^{2}}+0x-81\]
Now we will use the middle term to split
In middle term split apply on \[2\]
\[ax\text{+}bx+c\text{ }\], we produce ‘a’ by ‘c’ and then factor ‘ac’ in such a way that
if the product is ‘ac’ while the sum or difference is made up to ‘b’.
Now, we have middle term split on \[16{{x}^{2}}+0x-81\]
We have \[a=16,\,\,b=0\text{ and }c=-81~\]
So,
\[a\times c=16x-81=-1296~\]
Now we can see that \[36\times -36=1296\] and also \[36-36=0~\]\[\]
So, we use this to split the middle term.
So,
\[16{{x}^{2}}+0x-81=16{{x}^{2}}+\left( 36-36 \right)x-81~\]
Opening brackets, we get
\[~=16x{}^\text{2}+36x-36x-81~\]\[\]
We take common in the first \[2\] terms and the last \[2\] terms. So, we get –
\[~=4x\left( 4x+9 \right)-9\left( 4x+9 \right)~\]
As \[4x+9\] is same, so we get -
\[~=\left( 4x+9 \right)\left( 4x-9 \right)~\]
So, we get –
\[16x{}^\text{2}+0x-81=\left( 4x-9 \right)\left( 4x+9 \right)~\]
Simplifying we get factor as
\[16x{}^\text{2}-81=\left( 4x-9 \right)\left( 4x+9 \right)~\]

Note: While find the middle term using factor of $a\times c$, we need to keep in mind that when the sign of ‘a’ and ‘c’ are same then ‘b’ is obtained by addition only, if the sign of ‘a’ and ‘c’ are different then ‘b’ can be obtained using only subtraction. Key point to remember that degree of the equation will also tell us about the number of solutions also
\[2\] degree means given equation can have only \[2\] solution We can also use identity\[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\]to simple it faster \[16x{}^\text{2}-81\] as we know \[81={{9}^{2}}\text{ }and\text{ }16={{4}^{2}}\]
So \[16x{}^\text{2}-81\] can be written as \[{{4}^{2}}{{x}^{2}}-{{9}^{2}}~\]
Simply more we get
\[16{{x}^{2}}-81=\text{ }{{\left( 4x \right)}^{2}}-{{9}^{2}}~\]\[\]
Then using identity \[\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)~\]
\[16x{}^\text{2}-81=\left( 4x-9 \right)\left( 4x+9 \right)~\]