Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Our platform provides a seamless experience for finding reliable answers from a knowledgeable network of professionals. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
1. x = 1/ ( [tex] \sqrt{3} - \sqrt{2)} [/tex] = [tex] \sqrt{3}+ \sqrt{2} [/tex];
( [tex] \sqrt{x} -1/ \sqrt{x} )^{2} = x + 1/x - 2 =[/tex] =
( [tex] \sqrt{x} -1/ \sqrt{x} )^{2} = x + 1/x - 2 =[/tex] =
Answer with explanation:
Ques 1)
[tex]x=\dfrac{1}{\sqrt{3}-\sqrt{2}}[/tex]
Now we are asked to find the value of:
[tex]\sqrt{x}-\dfrac{1}{\sqrt{x}}[/tex]
We know that:
[tex](\sqrt{x}-\dfrac{1}{\sqrt{x}})^2=x+\dfrac{1}{x}-2[/tex]
Also:
[tex]x=\dfrac{1}{\sqrt{3}-\sqrt{2}}[/tex] could be written as:
[tex]x=\dfrac{1}{\sqrt{3}-\sqrt{2}}\times \dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\\\\\\x=\dfrac{\sqrt{3}+\sqrt{2}}{(\sqrt{3})^2-(\sqrt{2})^2}[/tex]
since, we know that:
[tex](a+b)(a-b)=a^2-b^2[/tex]
Hence,
[tex]x=\dfrac{\sqrt{3}+\sqrt{2}}{3-2}\\\\\\x=\sqrt{3}+\sqrt{2}[/tex]
Also,
[tex]\dfrac{1}{x}=\sqrt{3}-\sqrt{2}[/tex]
Hence, we get:
[tex](\sqrt{x}-\dfrac{1}{\sqrt{x}})^2=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}-2\\\\\\(\sqrt{x}-\dfrac{1}{\sqrt{x}})^2=2\sqrt{3}-2\\\\\\\sqrt{x}-\dfrac{1}{\sqrt{x}}=\sqrt{2\sqrt{3}-2}[/tex]
Hence,
[tex]\sqrt{x}-\dfrac{1}{\sqrt{x}}=\sqrt{2\sqrt{3}-2}[/tex]
Ques 2)
[tex]x=\dfrac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b}-\sqrt{a-2b}}[/tex]
on multiplying and dividing by conjugate of denominator we get:
[tex]x=\dfrac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b}-\sqrt{a-2b}}\times \dfrac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b}+\sqrt{a-2b}}\\\\\\x=\dfrac{(\sqrt{a+2b}+\sqrt{a-2b})^2}{(\sqrt{a+2b})^2-(\sqrt{a-2b})^2}\\\\\\x=\dfrac{(\sqrt{a+2b})^2+(\sqrt{a-2b})^2+2\sqrt{a+2b}\sqrt{a-2b}}{a+2b-a+2b}\\\\\\x=\dfrac{a+2b+a-2b+2\sqrt{a+2b}\sqrt{a-2b}}{4b}\\\\\\x=\dfrac{2a+2\sqrt{a^2-4b^2}}{4b}\\\\\\x^2=(\dfrac{2a+2\sqrt{a^2-4b^2}}{4b})^2\\\\\\x^2=\dfrac{(2a+2\sqrt{a^2-4b^2})^2}{16b^2}[/tex]
Hence, we have:
[tex]x^2=\dfrac{4a^2+4(a^2-4b^2)+8a\sqrt{a^2-4b^2}}{16b^2}\\\\\\x^2=\dfrac{4a^2+4a^2-16b^2+8a\sqrt{a^2-4b^2}}{16b^2}\\\\\\\\x^2=\dfrac{8a^2-16b^2+8a\sqrt{a^2-4b^2}}{16b^2}\\\\\\bx^2=\dfrac{8a^2-16b^2+8a\sqrt{a^2-4b^2}}{16b}\\\\\\bx^2=\dfrac{8a(a+\sqrt{a^2-4b^2})-16b^2}{16b}\\\\\\bx^2=\dfrac{8a(a+\sqrt{a^2-4b^2})}{16b}-\dfrac{16b^2}{16b}\\\\\\bx^2=\dfrac{a(a+\sqrt{a^2-4b^2})}{2b}-b\\\\\\bx^2=ax-b\\\\\\i.e.\\\\\\bx^2-ax+b=0[/tex]
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.
