ifx=1/2-√3,
find the value of x^3-2x^2-7x+5
Answers
Answer:
x=
2−
3
1
Multiply the numerator and denominator by the conjugate of the denominator
x=\frac{1}{2-\sqrt{3}}\cdot\frac{2+\sqrt{3}}{2+\sqrt{3}}=\frac{2+\sqrt{3}}{2^2-\sqrt{3}^2}=\frac{2+\sqrt{3}}{4-3}=2+\sqrt{3}x=
2−
3
1
⋅
2+
3
2+
3
=
2
2
−
3
2
2+
3
=
4−3
2+
3
=2+
3
\implies\,x^2=(2+\sqrt{3})^2=4+4\sqrt{3}+3=7+4\sqrt{3}⟹x
2
=(2+
3
)
2
=4+4
3
+3=7+4
3
Now consider the given expression
\begin{gathered}x^3-2x^2-7x+5\\=x^3-7x-2x^2+5\\=x(x^2-7)-2x^2+5\\=(2+\sqrt{3})(7+4\sqrt{3}-7)-2(7+4\sqrt{3})+5\\=(2+\sqrt{3})(4\sqrt{3})-14-8\sqrt{3}+5\\=8\sqrt{3}+12-14-8\sqrt{3}+5\\=\boxed{3}\end{gathered}
x
3
−2x
2
−7x+5
=x
3
−7x−2x
2
+5
=x(x
2
−7)−2x
2
+5
=(2+
3
)(7+4
3
−7)−2(7+4
3
)+5
=(2+
3
)(4
3
)−14−8
3
+5
=8
3
+12−14−8
3
+5
=
3
Answer:
answer is 3
Step-by-step explanation:
ur answer is 3
