Question

How do you solve \[{2^{2x}} + {2^{x + 2}} - 12 = 0\] ?

Answer 1

Hint: This question is a combination of indices and quadratic equations. We will use rules of indices and powers to modify the problem above. We will write the term with base 2 as we need to form a quadratic equation. Then we will use some substitution to solve the problem. Then we will find the roots or factors. Then we will reassign the substitution so used. This will give the solution to the problem above.

Complete step-by-step answer:
Given that,
 \[{2^{2x}} + {2^{x + 2}} - 12 = 0\]
We can write,
 \[{\left( {{2^x}} \right)^2} + {2^x}{2^2} - 12 = 0\] ……. \[{a^m}{a^n} = {a^{m + n}}\]
Now write the power of 2 in second term,
 \[{\left( {{2^x}} \right)^2} + 4\left( {{2^x}} \right) - 12 = 0\]
Now substitute \[{2^x} = s\] so that we will get the quadratic equation,
 \[{s^2} + 4s - 12 = 0\]
Now we will factorise the middle term such that the factors on adding will give the middle term and on multiplying will give the last term.
 \[{s^2} + 6s - 2s - 12 = 0\]
Now taking x common from first two terms and -2 common from last two terms.
 \[s\left( {s + 6} \right) - 2\left( {s + 6} \right) = 0\]
Now separating the brackets,
 \[\left( {s + 6} \right)\left( {s - 2} \right) = 0\]
Now separately assign the brackets to zero,
 \[s + 6 = 0ors - 2 = 0\]
 \[s = - 6ors = 2\]
Now -6 is not the correct factor here because it is the answer for power.
 \[s = 2\] is the correct factor we will proceed with.
Now we will substitute the value,
 \[s = {2^x}\]
 \[
  2 = {2^x} \\
  {2^1} = {2^x} \;
\]
Thus since the bases are same we can equate the powers directly,
 \[x = 1\] .
This is the correct answer.
So, the correct answer is “ \[x = 1\] .”.

Note: Note that initially don’t get confused with the third term that is 12 like it is not any power if 2. So there is the key point that the question is of indices and quadratic equations. Also note that we have used substitution for the equation so don’t forget to substitute the value for the correct answer.