Question

if root under x subtract 1/root under 5 = root under 5 then the value of x is​

Answer 1

Answer:

Answer:

x² + y² = 7

Step-by-step explanation:

Give values of x and y are

x=\frac{\sqrt{5}+1}{\sqrt{5}-1}\:\:and\:\:y=\frac{\sqrt{5}-1}{\sqrt{5}+1}x=5−15+1andy=5+15−1

To find: x² + y²

first we find value of x and y by rationalizing the denominator

Consider,

x=\frac{\sqrt{5}+1}{\sqrt{5}-1}x=5−15+1

x=\frac{\sqrt{5}+1}{\sqrt{5}-1}\times\frac{\sqrt{5}+1}{\sqrt{5}+1}x=5−15+1×5+15+1

x=\frac{(\sqrt{5}+1)^2}{(\sqrt{5}-1)(\sqrt{5}+1)}x=(5−1)(5+1)(5+1)2

x=\frac{(\sqrt{5})^2+1+2\times\sqrt{5}}{(\sqrt{5})^2-1}x=(5)2−1(5)2+1+2×5

x=\frac{5+1+2\times\sqrt{5}}{5-1}x=5−15+1+2×5

x=\frac{6+2\times\sqrt{5}}{4}x=46+2×5

x=\frac{3+\sqrt{5}}{2}x=23+5

y=\frac{\sqrt{5}-1}{\sqrt{5}+1}y=5+15−1

x=\frac{\sqrt{5}-1}{\sqrt{5}+1}\times\frac{\sqrt{5}-1}{\sqrt{5}-1}x=5+15−1×5−15−1

x=\frac{(\sqrt{5}-1)^2}{(\sqrt{5}+1)(\sqrt{5}-1)}x=(5+1)(5−1)(5−1)2

x=\frac{(\sqrt{5})^2+1-2\times\sqrt{5}}{(\sqrt{5})^2-1}x=(5)2−1(5)2+1−2×5

x=\frac{5+1-2\times\sqrt{5}}{5-1}x=5−15+1−2×5