If x = √+3 +√−3 √+3 −√−3 , Prove that 3ax2 – 2bx + 3a = 0
Answers
Answer:
Step-by-step explanation:Given:- x=
a+3b
−
a−3b
a+3b
+
a−3b
To prove:- 3bx
2
−2ax+3b=0
Proof:-
x=
a+3b
−
a−3b
a+3b
+
a−3b
Multiply and divide the above equation by conjugate of its denominator, we have
x=
a+3b
−
a−3b
a+3b
+
a−3b
×
a+3b
+
a−3b
a+3b
+
a−3b
x=
(
a+3b
)
2
−(
a−3b
)
2
(
a+3b
+
a−3b
)
2
x=
(a+3b)−(a−3b)
(a+3b)+(a−3b)+2
(a+3b)(a−3b)
x=
a+3b−a+3b
a+3b+a−3b+2
(a+3b)(a−3b)
⇒x=
6b
2a+2
a
2
−(3b)
2
⇒6bx=2a+2
a
2
−9b
2
⇒3bx−a=
a
2
−9b
2
Squring both sides, we have
(3bx−a)
2
=(
a
2
−9b
2
)
2
9b
2
x
2
+a
2
−6abx=a
2
−9b
2
9b
2
x
2
−6abx+9b
2
=0
⇒3b(3bx
2
−2ax+3b)=0
⇒3bx
2
−2ax+3b=0
Hence proved.