
Let x be real number such that $ {x^3} + 4x = 8 $ , then the value of $ {x^7} + 64{x^2} $ is
A.136
B.146
C.128
D.156
Answer
569.4k+ views
Hint: There can be several ways to solve a polynomial equation. You can use any of them but first, you would have to think of establishing a link between the given equation and the equation that you have to find. Once you find that out, you can substitute the given values whenever needed and find out the correct answer.
Complete step-by-step answer:
This question can be solved by two methods,
First approach –
We are given that $ {x^3} + 4x = 8 $
On squaring both sides, we get –
$
\Rightarrow {({x^3} + 4x)^2} = {(8)^2} \\
\Rightarrow {x^6} + 16{x^2} + 8{x^4} = 64 \;
$
To make the above equation of degree 7, we multiply x on both sides,
$\Rightarrow {x^7} + 16{x^3} + 8{x^5} = 64x $
Add $ 16{x^3} $ on both sides
$ {x^7} + 32{x^3} + 8{x^5} = 64x + 16{x^3} $
Now we take $ 8{x^2} $ common on the left-hand side and $ 16 $ common on the right-hand side.
$ {x^7} + 8{x^2}({x^3} + 4x) = 16({x^3} + 4x) $
We know the value of $ {x^3} + 4x $ , substituting the value in the above equation –
$
\Rightarrow {x^7} + 8{x^2}(8) = 16(8) \\
\Rightarrow {x^7} + 64{x^2} = 128 \;
$
Thus, we have got the required answer.
Second approach –
We have to find $ {x^7} + 64{x^2} $
$ {x^7} $ can be written as $ x{({x^3})^2} $ , so the above equation becomes -
$ x \times {({x^3})^2} + 64{x^2} $
We are given that,
$
\Rightarrow {x^3} + 4x = 8 \\
\Rightarrow {x^3} = 8 - 4x \;
$
Using this value in the equation that we have to find
$
\Rightarrow x{(8 - 4x)^2} + 64{x^2} \\
= x(64 + 16{x^2} - 64x) + 64{x^2} \\
= 64x + 16{x^3} - 64{x^2} + 64{x^2} \\
= 16{x^3} + 64x \\
= 16({x^3} + 4x) \\
= 16(8) \\
= 128 \;
$
That is, $ {x^7} + 64{x^2} = 128 $
So, the correct answer is “Option C”.
Note: An expression composed of variables, constants and exponents, that contain only the operations of subtraction, addition, multiplication and division but there cannot be a variable in the denominator. The highest exponent in the polynomial equation is called the degree of that equation. While taking commons, keep the given values in mind so that after taking common, they can be easily substituted.
Complete step-by-step answer:
This question can be solved by two methods,
First approach –
We are given that $ {x^3} + 4x = 8 $
On squaring both sides, we get –
$
\Rightarrow {({x^3} + 4x)^2} = {(8)^2} \\
\Rightarrow {x^6} + 16{x^2} + 8{x^4} = 64 \;
$
To make the above equation of degree 7, we multiply x on both sides,
$\Rightarrow {x^7} + 16{x^3} + 8{x^5} = 64x $
Add $ 16{x^3} $ on both sides
$ {x^7} + 32{x^3} + 8{x^5} = 64x + 16{x^3} $
Now we take $ 8{x^2} $ common on the left-hand side and $ 16 $ common on the right-hand side.
$ {x^7} + 8{x^2}({x^3} + 4x) = 16({x^3} + 4x) $
We know the value of $ {x^3} + 4x $ , substituting the value in the above equation –
$
\Rightarrow {x^7} + 8{x^2}(8) = 16(8) \\
\Rightarrow {x^7} + 64{x^2} = 128 \;
$
Thus, we have got the required answer.
Second approach –
We have to find $ {x^7} + 64{x^2} $
$ {x^7} $ can be written as $ x{({x^3})^2} $ , so the above equation becomes -
$ x \times {({x^3})^2} + 64{x^2} $
We are given that,
$
\Rightarrow {x^3} + 4x = 8 \\
\Rightarrow {x^3} = 8 - 4x \;
$
Using this value in the equation that we have to find
$
\Rightarrow x{(8 - 4x)^2} + 64{x^2} \\
= x(64 + 16{x^2} - 64x) + 64{x^2} \\
= 64x + 16{x^3} - 64{x^2} + 64{x^2} \\
= 16{x^3} + 64x \\
= 16({x^3} + 4x) \\
= 16(8) \\
= 128 \;
$
That is, $ {x^7} + 64{x^2} = 128 $
So, the correct answer is “Option C”.
Note: An expression composed of variables, constants and exponents, that contain only the operations of subtraction, addition, multiplication and division but there cannot be a variable in the denominator. The highest exponent in the polynomial equation is called the degree of that equation. While taking commons, keep the given values in mind so that after taking common, they can be easily substituted.
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