Question 353546
Q1.IF {{{x^3+1/x^3=2}}} then find the value of x+1/x
Let's start by eliminating the fraction. Multiply both sides by {{{x^3}}}:
{{{x^6 + 1 = 2x^3}}}
To solve a non-linear equation, we often get one side equal to zero and factor:
{{{x^6 - 2x^3 + 1 = 0}}}
Solving this equation is much easier if we notice that the exponent of 6 is twice the exponent of 3. That makes this equation in quadratic form for {{{x^3}}}. Looking at the expression this way we can, perhaps, see that the expression fits the pattern: {{{a^2 + 2ab + b^2}}} which we know is, in factored form: {{{(a+b)^2}}}. So you equation factors into:
{{{x^3 - 1)^2 = 0}}}
By the Zero Product Property we know that this (or any) product is zero <i>only</i> if one of the factors is zero. So:
{{{x^3 - 1 = 0}}}
Solving this we get x = 1.
This makes x + 1/x = 1 + 1/1 = 1 + 1 = 2.<br>
Q2.FIND THE VALUE OF {{{(x+y)^3+(x-2y+2)}}}
This problem does not make sense. You can't find a value unless you have an equation and this has no equals sign. The only thing you can do with this expression is simplify it. That means cubing (x+y) correctly (Hint: It is <b>not</b> {{{x^3 + y^3}}}!) and then combining like terms, if any.<br>
Q3.IF {{{(x+1/x)*(x-1/x)*(x^2+1/x^2)*(x^4+1/x^4)*(x^8+1/x^8)=3/2}}} .then find the value of {{{x^16}}}
The first two factors{{{(x+1/x)*(x-1/x)}}} fit the pattern of (a+b)(a-b) which we know from the pattern to be equal to {{{a^2 - b^2}}}. So your first two factors are equal to {{{x^2 - 1/x^2}}}. But this and the third factor fit the same pattern again. So the first three factors are equal to {{{x^4 - 1/x^4}}}. But this, combined with the next factor fit the pattern again. And this repeats once more time making the entire left side of the equation equal to:
{{{x^16 - 1/x^16}}}. So now our equation looks like:
{{{x^16 - 1/x^16 = 3/2}}}
We can solve this just like we did problem 1. Get rid of the fraction:
{{{x^32 - 1 = (3/2)x^16}}}
Get one side equal to zero:
{{{x^32 - (3/2)x^16 - 1 = 0}}}
This, like problem 1, is an equation in "quadratic form". This time it is in quadratic form for {{{x^16}}}. This equation, unlike problem 1, does not factor easily. When we can't factor, we resort to the Quadratic Formula. This gives us:
{{{x^16 = (-(-3/2) +- sqrt((-3/2)^2 - 4(1)(-1)))/2(1)}}}
Simplifying.
{{{x^16 = (-(-3/2) +- sqrt(9/4 - 4(1)(-1)))/2(1)}}}
{{{x^16 = (3/2 +- sqrt(9/4 + 4))/2}}}
{{{x^16 = (3/2 +- sqrt(9/4 + 16/4))/2}}}
{{{x^16 = (3/2 +- sqrt(25/4))/2}}}
{{{x^16 = (3/2 +- 5/2)/2}}}
In long form this is:
{{{x^16 = (3/2 + 5/2)/2}}} or {{{x^16 = (3/2 - 5/2)/2}}}
Looking at the second equation, we see that the fraction will be negative. And since {{{x^16}}} cannot be equal to a negative number, there will be no solutions to the second equation. So we only have to solve the first equation. we continue by simplifying:
{{{x^16 = (8/2)/2}}}
{{{x^16 = 4/2}}}
{{{x^16 = 2}}}