Question

Find the value of x if (97-x)^1/4 + x^1/4 = 5...

Answer 1

$(97 - x)^\frac{1}{4} + x^\frac{1}{4} = 5$

Square both sides:

$[ (97 - x)^\frac{1}{4} + x^\frac{1}{4}]^2 = 5^2$

$(97 - x)^\frac{1}{2} + x^\frac{1}{2} + 2(97 - x)^\frac{1}{4}x^\frac{1}{4}=25$

Subtract "2ab" from both sides:

$(97 - x)^\frac{1}{2} + x^\frac{1}{2} = 25 - 2[x (97 - x)]^\frac{1}{4}$

Square both sides:

$[(97 - x)^\frac{1}{2} + x^\frac{1}{2}]^2 = [25 - 2[x (97 - x)]^\frac{1}{4}]^2$

$(97 - x) + x + 2[ x(97 - x)]^\frac{1}{2} = 625 + 4[x{97 - x)]^\frac{1}{2} - 100[x (97 - x)]^\frac{1}{4}$

Simplify:

$528 + 2[x{97 - x)]^\frac{1}{2} - 100[x (97 - x)]^\frac{1}{4} = 0$

Let u = [ x(97 - x)] ^1/4:

$528 + 2u^2 - 100u = 0$

$2u^2 - 100u + 528 = 0$

$u^2 - 50u + 264 = 0$

$(u - 6)(u - 44) = 0$

$u = 6 \ or \ u = 44$

Solve x:

$\text {When u = 6}$

$[ x(97 - x)] ^\frac{1}{4} = 6$

$97x - x^2 = 1296$

$x^2 - 97x + 1296 = 0$

$(x - 16) (x - 81) = 0$

$x = 16 \ or \ x = 81$

$\text {When u = 44}$

$[ x(97 - x)] ^\frac{1}{4} = 44$

$[ x(97 - x)]= 3748096$

$x^2 - 97x + 3748096$

⇒ No solution

Answer: x = 16 or x = 81

Answer 2

Hey there!

Rewrite this following equation with respect to a variable that is, "u".

$\bf{u = (97 - x)^{\frac{1}{4}}}$ AND for "x",  $\bf{x = - u^4 + 97}$.

Therefore we get,

$\bf{u + (- u^4 + 97)^{\frac{1}{4}}) = 5}$

Solve this whole equation!

By subtracting "u' from both the sides:

$\bf{u + (- u^4 + 97)^{\frac{1}{4}}) - u = 5 - u}$

$\bf{(- u^4 + 97)^{\frac{1}{4}}) = 5 - u}$

Taking both the sides of equation as per the exponential power of "4".

$\bf{((- u^4 + 97)^{\frac{1}{4}})^4 = (5 - u)^4}$

Apply the rule of exponents that is, $\bf{(a^b)^c = a^{b \times c}}$.

$\bf{(- u^4 + 97)^{\frac{1}{4} \times 4} = (5 - u)^4}$

$\bf{(- u^4 + 97) = (5 - u)^4}$

By applying the principles and basics of Binomial theorem that is:

$\boxed{\bf{(a + b)^n = \textstyle\sum_{i = 0}^n \binom{n}{i} a^{(n - i)} b^i}}$

Here, a = 5 and b = - u.

$\bf{= \textstyle\sum_{i = 0}^4 \binom{4}{i} \times 5^{(4 - i)} (- u)^i}$

Now, Expand this given summation (I'll solve this in parts for better understanding), for i = 0, 1, 2, 3 and 4, respectively :

$\bf{= \frac{4!}{0! (4 - 0)!} \times 5^4 (- u)^0}$

+  $\bf{\frac{4!}{1! (4 - 1)!} \times 5^3 (- u)^1}$

+   $\bf{\frac{4!}{2! (4 - 2)!} \times 5^2 (- u)^2}$

+  $\bf{\frac{4!}{3! (4 - 3)!} \times 5^1 (- u)^3}$

+  $\bf{\frac{4!}{4! (4 - 4)! \times 5^0 (- u)^4}$

Individually solving for the required summed values of "i" ;

For i = 0,

$\bf{\frac{4!}{0! (4 - 0)!} \times 5^4 (- u)^0}$

$\bf{\frac{5^4 \times 4!}{4!}}$

$\bf{625}$

For i = 1,

$\bf{\frac{4!}{1! (4 - 1)!} \times 5^3 (- u)^1}$

$\bf{- \frac{5^3 \times 4!}{1! (4 - 1)!} \times u}$

$\bf{- \frac{5^3 \times 4}{1!} \times u}$

$\bf{- 500u}$

For i = 2,

$\bf{\frac{4!}{2! (4 - 2)!} \times 5^2 (- u)^2}$

$\bf{\frac{5^2 \times 4 \times 3 u^2}{2!}}$

$\bf{\frac{300u^2}{2!}}$

$\bf{150u^2}$

For i = 3,

$\bf{\frac{4!}{3! (4 - 3)!} \times 5^1 (- u)^3}$

$\bf{- \frac{5 \times 4! u^3}{3! \times 1!}}$

$\bf{- \frac{20u^3}{1}}$

$\bf{- 20u^3}$

For i = 4,

$\bf{\frac{4!}{4! (4 - 4)! \times 5^0 (- u)^4}$

$\bf{\frac{(- u)^4}{(4 - 4)}!}$

$\bf{u^4}$

We get,

$\bf{- u^4 + 97 = 625 - 500u + 150u^2 - 20u^3 + u^4}$

Solve this whole equation :

Switch both the sides, subtract both sides by a value of "97", Simplify and add $\bf{u^4}$ on both sides and simplify to obtain :

$\bf{2u^4 - 20u^3 + 150u^2 - 500u + 528 = 0}$

Solve by the method of factoring the terms :

Factor out the common term "2" :

$\bf{2(u^4 - 10u^3 + 75u^2 - 250u + 264 = 0}$

Factor the inner brackets and apply rational root theorem that is,

$\bf{a_0 = 264 \: \: \: a_n = 1}$

Following are the divisors of $\bf{a_0}$ : 1, 2, 3, 33, 4, 66, 6, 8, 11, 12, 132, 44, 22, 24, 264, 88. And divisor for $\bf{a_n}$ : 1.

Now, check the following rational numbers:

$\bf{+-\frac{1, 2, 3, 33, 4, 66, 6, 8, 11, 12, 132, 44, 22, 24, 264, 88}{1}}$

Therefore, $\bf{2}{1}}$ is a root of the expression so, factoring out (u -2) :

$\bf{= 2(u - 2)(u^3 - 8u + 59u - 132)}$

Apply the same rational root theorem for the inner brackets to get the expression :

$\bf{a_0 = 132 \: \: \: a_n = 1}$

Following are the divisors of $\bf{a_0}$ : 1, 2, 33, 3, 4, 66, 6, 22, 11, 12, 132, 44. And divisor of $\bf{a-0}$ : 1.

Now, check the following rational numbers:

$\bf{+- \frac{1, 2, 33, 3, 4, 66, 6, 22, 11, 12, 132, 44}{1}}$

Therefore, $\bf{3}{1}}$ is a root of the expression so, factoring out (u - 3) :

$\bf{= 2(u - 2)(u - 3)(u^2 - 5u + 44)}$

Now, by applying the zero factor principle into this and solving for "u" ;

Solve for "u - 2" :

= u - 2 = 0

$\boxed{\bf{u = 2}}$

Solve for "u - 3" :

= u - 3 = 0

$\boxed{\bf{u = 3}}$

Solve for $\bf{u^2 - 5u + 44 = 0}$ :

For a specific quadratic equation there should be a discriminant of $\bf{b^2 - 4ac}$ for an equation of $\bf{ax^2 + bx + c = 0}$ :

$\bf{(- 5)^2 - 4 \times 1 \times 44}$

$\bf{= - 151}$

Discriminant can't be in a negative form for a set solution of "u", therefore there's no final solution for third part.

The final solutions after reaching different values for "u" is :

$\boxed{\huge{u = 2 \: \: \: u = 3}}$

Moving on to obtain the final values for the variable "x" :

Since, $\bf{u = (97 - x)^{\frac{1}{4}}}$

Here, "u = 2" and "u = 3" :

Solving for "u = 2" :

Powering the expressions to the value of "4" :

$\bf{((97 - x)^{\frac{1}{4}})^4 = 2^4}$

$\bf{97 - x = 16}$

$\boxed{\huge{x = 81}}$

[Character limit] Similarly find the second value of "x" by solving "u =3".

The two values are : x = 81, x = 16.