Answer
Verified
460.8k+ views
Hint: In this question take ${2^x} = y$ which will help you to obtain a new quadratic equation in terms of y to find the roots of the new quadratic equation using the factoring method, using the value of roots you can obtain the value of x and easier to identify the correct option.
Complete step-by-step solution:
According to the given information we have equation ${2^{2x + 3}} - 57 = 65\left( {{2^x} - 1} \right)$
Since we know that ${a^{n + m}} = {a^n} \times {a^m}$
Therefore, the above equation can be re-written as; ${\left( {{2^x}} \right)^2} \times {2^3} = 65\left( {{2^x} - 1} \right) + 57$................ (equation 1)
let ${2^x} = y$ .......... (equation 2)
substituting the y in the equation 1 we get
${y^2} \times {2^3} = 65\left( {y - 1} \right) + 57$
$ \Rightarrow $$8{y^2} = 65y - 65 + 57$
$ \Rightarrow $$8{y^2} - 65y + 8 = 0$
Now using the factoring method to find the roots of the above quadratic equation we get
$8{y^2} - \left( {64 + 1} \right)y + 8 = 0$
$ \Rightarrow $$8{y^2} - \left( {64 + 1} \right)y + 8 = 0$
$ \Rightarrow $$8{y^2} - 64y - 1y + 8 = 0$
$ \Rightarrow $$8y\left( {y - 8} \right) - 1\left( {y - 8} \right) = 0$
$ \Rightarrow $$\left( {8y - 1} \right)\left( {y - 8} \right) = 0$
So, $y = 8$ or $y = \dfrac{1}{8}$
Now substituting the value of y in the equation 2
For y = 8
$\Rightarrow {2^x} = 8$
$\Rightarrow {2^x} = {\left( 2 \right)^3}$
Therefore, x = 3
Now for $y = \dfrac{1}{8}$
$\Rightarrow {2^x} = \dfrac{1}{8}$
$\Rightarrow {2^x} = {\left( 2 \right)^{ - 3}}$
Therefore, $x = – 3 $
So, the roots of the given is $ \pm 3$
Hence, option B is the correct option.
Note: In the above solution we used a term “quadratic equation” which can be defined as the polynomial equation of degree two which implies that this equation has only one term with highest power two. A quadratic equation general form is represented as $a{x^2} + bx + c = 0$ here x is variable which is unknown and a, b, c are the numerical coefficients. The unknown values of x can be find using the quadratic formula which is given as; \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
Complete step-by-step solution:
According to the given information we have equation ${2^{2x + 3}} - 57 = 65\left( {{2^x} - 1} \right)$
Since we know that ${a^{n + m}} = {a^n} \times {a^m}$
Therefore, the above equation can be re-written as; ${\left( {{2^x}} \right)^2} \times {2^3} = 65\left( {{2^x} - 1} \right) + 57$................ (equation 1)
let ${2^x} = y$ .......... (equation 2)
substituting the y in the equation 1 we get
${y^2} \times {2^3} = 65\left( {y - 1} \right) + 57$
$ \Rightarrow $$8{y^2} = 65y - 65 + 57$
$ \Rightarrow $$8{y^2} - 65y + 8 = 0$
Now using the factoring method to find the roots of the above quadratic equation we get
$8{y^2} - \left( {64 + 1} \right)y + 8 = 0$
$ \Rightarrow $$8{y^2} - \left( {64 + 1} \right)y + 8 = 0$
$ \Rightarrow $$8{y^2} - 64y - 1y + 8 = 0$
$ \Rightarrow $$8y\left( {y - 8} \right) - 1\left( {y - 8} \right) = 0$
$ \Rightarrow $$\left( {8y - 1} \right)\left( {y - 8} \right) = 0$
So, $y = 8$ or $y = \dfrac{1}{8}$
Now substituting the value of y in the equation 2
For y = 8
$\Rightarrow {2^x} = 8$
$\Rightarrow {2^x} = {\left( 2 \right)^3}$
Therefore, x = 3
Now for $y = \dfrac{1}{8}$
$\Rightarrow {2^x} = \dfrac{1}{8}$
$\Rightarrow {2^x} = {\left( 2 \right)^{ - 3}}$
Therefore, $x = – 3 $
So, the roots of the given is $ \pm 3$
Hence, option B is the correct option.
Note: In the above solution we used a term “quadratic equation” which can be defined as the polynomial equation of degree two which implies that this equation has only one term with highest power two. A quadratic equation general form is represented as $a{x^2} + bx + c = 0$ here x is variable which is unknown and a, b, c are the numerical coefficients. The unknown values of x can be find using the quadratic formula which is given as; \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
Recently Updated Pages
Fill in the blanks with suitable prepositions Break class 10 english CBSE
Fill in the blanks with suitable articles Tribune is class 10 english CBSE
Rearrange the following words and phrases to form a class 10 english CBSE
Select the opposite of the given word Permit aGive class 10 english CBSE
Fill in the blank with the most appropriate option class 10 english CBSE
Some places have oneline notices Which option is a class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
Which are the Top 10 Largest Countries of the World?
What is the definite integral of zero a constant b class 12 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Define the term system surroundings open system closed class 11 chemistry CBSE
Full Form of IASDMIPSIFSIRSPOLICE class 7 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE