X=√(p +2q) -. √(p-2q) / √(p+2q)+√(p+2q) prove qx^2-px-q
Answers
Step-by-step explanation:
From given x=\frac{\sqrt{p+2q}+\sqrt{p-2q}}{\sqrt{p+2q}-\sqrt{p-2q}}x=
p+2q
−
p−2q
p+2q
+
p−2q
we have proved that qx^2-px+q=0qx
2
−px+q=0
Step-by-step explanation:
Given that x=\frac{\sqrt{p+2q}+\sqrt{p-2q}}{\sqrt{p+2q}-\sqrt{p-2q}}x=
p+2q
−
p−2q
p+2q
+
p−2q
To prove that qx^2-px+q=0qx
2
−px+q=0
x=\frac{\sqrt{p+2q}+\sqrt{p-2q}}{\sqrt{p+2q}-\sqrt{p-2q}}x=
p+2q
−
p−2q
p+2q
+
p−2q
Multiply and dividing by the conjugate \sqrt{p+2q}+\sqrt{p-2q}
p+2q
+
p−2q
we get
x=\frac{\sqrt{p+2q}+\sqrt{p-2q}}{\sqrt{p+2q}-\sqrt{p-2q}}\times \frac{\sqrt{p+2q}+\sqrt{p-2q}}{\sqrt{p+2q}+\sqrt{p-2q}}x=
p+2q
−
p−2q
p+2q
+
p−2q
×
p+2q
+
p−2q
p+2q
+
p−2q
x=\frac{(\sqrt{p+2q}+\sqrt{p-2q})^2}{\sqrt{p+2q}^2-\sqrt{p-2q}^2}x=
p+2q
2
−
p−2q
2
(
p+2q
+
p−2q
)
2
( using the properties (a+b)^2=a^2+b^2+2ab(a+b)
2
=a
2
+b
2
+2ab and (a-b)(a+b)=a^2-b^2(a−b)(a+b)=a
2
−b
2
)
x=\frac{\sqrt{p+2q}^2+\sqrt{p-2q}^2+2\sqrt{p+2q}\sqrt{p-2q}}{p+2q-(p-2q)}x=
p+2q−(p−2q)
p+2q
2
+
p−2q
2
+2
p+2q
p−2q
x=\frac{p+2q+p-2q+2\sqrt{(p+2q)(p-2q)}}{p+2q-p+2q}x=
p+2q−p+2q
p+2q+p−2q+2
(p+2q)(p−2q)
x=\frac{2p+2\sqrt{p^2-(2q)^2}}{4q}x=
4q
2p+2
p
2
−(2q)
2
( using the property (a-b)(a+b)=a^2-b^2(a−b)(a+b)=a
2
−b
2
)
x=2(\frac{p+\sqrt{p^2-4q^2}}{4q})x=2(
4q
p+
p
2
−4q
2
)
x=\frac{p+\sqrt{p^2-4q^2}}{2q}x=
2q
p+
p
2
−4q
2
(2q)x=p+\sqrt{p^2-4q^2}(2q)x=p+
p
2
−4q
2
2qx-p=\sqrt{p^2-4q^2}2qx−p=
p
2
−4q
2
Now squaring on both sides we get
(2qx-p)^2=(\sqrt{p^2-4q^2})^2(2qx−p)
2
=(
p
2
−4q
2
)
2
(2qx)^2+p^2-2(2qx)(p)=p^2-4q^2(2qx)
2
+p
2
−2(2qx)(p)=p
2
−4q
2
4q^2x^2-4qpx=-4q^24q
2
x
2
−4qpx=−4q
2
Dividing by 4q on both sides we get
\frac{4q^2x^2-4qpx}{4q}=-\frac{4q^2}{4q}
4q
4q
2
x
2
−4qpx
=−
4q
4q
2
qx^2-px=-qqx
2
−px=−q
Therefore qx^2-px+q=0qx
2
−px+q=0
Hence proved
Answer:
In attachment I have answer . ↑↑↑^
