Question

if X equal to 1 by 2 minus root 3 then the value of x square minus one by X is 14​

Answer 1

Proving:

Step-by-step explanation:

$\ Given \ value:\\\\x=\frac{1}{2-\sqrt{3}} \ \ \ \ \ \ and \ \ \ \ x^2- \frac{1}{x} =14\\\\\ Find:\\\\x= ?\\\\\ Solution: \\\\x=\frac{1}{2-\sqrt{3}} \times \frac{2+\sqrt{3}}{2+\sqrt{3}} \\\\x=\frac{2+\sqrt{3}}{(2)^2-(\sqrt{3})^2} \\\\x=\frac{2+\sqrt{3}}{4-3} \\\\x= 2+\sqrt{3}\\\\x^2= (2+\sqrt{3})^2\\\\x^2 = 4+3+4\sqrt{3}\\\\\ x^2 = 7+4\sqrt{3}\\\\\ put \ the \ value \ of \ x \ and \ x^2 \ in \ second \ equation \\\\\ Equation: \\\\x^2- \frac{1}{x} =14$

$(7+4\sqrt{3}) + \frac{1}{2+\sqrt{3}} =14\\\\\frac{(7+4\sqrt{3})(2+\sqrt{3})+1}{2+\sqrt{3}} =14\\\\\frac{(14+7\sqrt{3}+8\sqrt{3}+12)+1}{2+\sqrt{3}} =14\\\\\frac{27+15\sqrt{3}}{2+\sqrt{3}} =14\\\\\frac{27+15\sqrt{3})}{2+\sqrt{3}} =14\\\\27+15\sqrt{3} =14(2+\sqrt{3})\\\\27+15\sqrt{3} =28+14\sqrt{3}\\\\15\sqrt{3}-14\sqrt{3} =28-27\\\\\sqrt{3}=1$

$\ L.H.S \neq R.H.S \\\ So, x =\frac{1}{2-\sqrt{3}} \ is \ not \ solution \ to \ this \ question$

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• Proving: https://brainly.in/question/16092352