Solve for x :
[tex] \star \: \boxed{\: \sf{4 \bigg( x - \dfrac{1}{x} \bigg) ^{2} - 4 \bigg(x + \dfrac{1}{x} \bigg) + 1 = 0}}[/tex]
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Solve for x :
[tex] \star \: \boxed{\: \sf{4 \bigg( x - \dfrac{1}{x} \bigg) ^{2} - 4 \bigg(x + \dfrac{1}{x} \bigg) + 1 = 0}}[/tex]
» Please Help!
» No Spamming!
» Thankss!
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Answer:
It appears that you have two equations with different notations, but I'll assume they are equivalent and solve for x in the second equation:
4(x - x1)² - 4(x + x1) + 1 = 0
Now, let's simplify and solve for x:
4x² - 8x1x + 4x1² - 4x - 4x1 + 1 = 0
Combine like terms:
4x² - (8x1x + 4x + 4x1) + 4x1² + 1 = 0
Now, the equation is in the form of a quadratic equation (ax² + bx + c = 0), where:
a = 4
b = - (8x1x + 4x + 4x1)
c = 4x1² + 1
You can use the quadratic formula to solve for x:
x = (-b ± √(b² - 4ac)) / (2a)
Substitute the values of a, b, and c:
x = (-(8x1x + 4x + 4x1) ± √((8x1x + 4x + 4x1)² - 4 * 4 * (4x1² + 1))) / (2 * 4)
Now, you can simplify this expression to find the solutions for x. Please note that these solutions may involve complex numbers depending on the values of x1 and other constants in your problem.
Explanation:
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Verified answer
[tex]\boxed{{(a+b)}^{2}={a}^{2}+2ab+{b}^{2}}[/tex]
We want [tex]{x}^{2}+\dfrac{1}{{x}^{2}}[/tex] in terms of lower powers.
Set [tex]u=x+\dfrac{1}{x}[/tex].
[tex]\therefore{x}^{2}+\dfrac{1}{{x}^{2}}={u}^{2}-2[/tex]
[tex]4{\bigg(x-\dfrac{1}{x}\bigg)}^{2}-4\bigg(x+\dfrac{1}{x}\bigg)+1=0[/tex]
[tex]\iff4({u}^{2}-4)-4u+1=0[/tex]
[tex]\iff4{u}^{2}-4u-15=0[/tex]
[tex]\iff(2u+3)(2u-5)=0[/tex]
[tex]\iff 2u+3=0\textrm{ or }2u-5=0[/tex]
[tex]\iff 2\bigg(x+\dfrac{1}{x}\bigg)+3=0\textrm{ or }2\bigg(x+\dfrac{1}{x}\bigg)-5=0[/tex]
[tex]\iff 2{x}^{2}+3x+2=0\textrm{ or }2{x}^{2}-5x+2=0[/tex]
[tex]\iff x=\dfrac{-3\pm\sqrt{7}i}{4}\textrm{ or }x=\dfrac{5\pm3}{4}[/tex]
[tex]\iff \boxed{x=\dfrac{-3\pm\sqrt{7}i}{4}\textrm{ or }x=2\textrm{ or }x=\dfrac{1}{2}}[/tex]