If x = √7+ √3 and xy = 4, then x4 + y4 =
Answers
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If x^4-79x^2+1=0 then find the value of x^3+1/x^3
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Step-by-step explanation:
Given:
x=\sqrt{7}+\sqrt{3}\;\text{and}\;xy=4x=
7
+
3
andxy=4
\textbf{To find:}To find:
\text{The value of $x^4+y^4$}The value of x
4
+y
4
\text{consider,}consider,
xy=4xy=4
y=\frac{4}{x}y=
x
4
\text{To rationalize the denominator}To rationalize the denominator
y=\frac{4}{\sqrt{7}+\sqrt{3}}{\times}\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}-\sqrt{3}}y=
7
+
3
4
×
7
−
3
7
−
3
y=4(\frac{\sqrt{7}-\sqrt{3}}{7-3})y=4(
7−3
7
−
3
)
\implies\bf\;y=\sqrt{7}-\sqrt{3}⟹y=
7
−
3
\text{Using the identity,}Using the identity,
\boxed{\bf\;a^2+b^2=(a+b)^2-2ab}
a
2
+b
2
=(a+b)
2
−2ab
\text{we get}we get
x^4+y4=(x^2+y^2)^2-2x^2y^2x
4
+y4=(x
2
+y
2
)
2
−2x
2
y
2
x^4+y4=((\sqrt{7}+\sqrt{3})^2+(\sqrt{7}-\sqrt{3})^2)^2-2(xy)^2x
4
+y4=((
7
+
3
)
2
+(
7
−
3
)
2
)
2
−2(xy)
2
x^4+y4=(7+3+2\sqrt{7}\sqrt{3}+7+3-2\sqrt{7}\sqrt{3})^2-2(4)^2x
4
+y4=(7+3+2
7
3
+7+3−2
7
3
)
2
−2(4)
2
x^4+y4=(20)^2-32x
4
+y4=(20)
2
−32
x^4+y4=400-32x
4
+y4=400−32
\implies\boxed{\bf\;x^4+y4=368}⟹
x
4
+y4=368
\therefore\textbf{The value of $\bf\;x^4+y^4$ is 368}∴The value of x
4
+y
4
is 368