Solve the following quadratic equations by factorization: ax² +(4a² -3b)x-12ab=0
Answers
SOLUTION :
Given : ax² + (4a² - 3b)x -12ab = 0
ax² + 4a²x - 3bx -12ab = 0
ax(x + 4a) – 3b(x + 4a) = 0
(x + 4a)(ax - 3b) = 0
x + 4a = 0 or ax - 3b = 0
x = - 4a or ax = 3b
x = - 4a or x = 3b/a
Hence, the roots of the quadratic equation ax² + (4a² - 3b)x -12ab = 0 are - 4a or 3b/a
★★ METHOD TO FIND SOLUTION OF a quadratic equation by FACTORIZATION METHOD :
We first write the given quadratic polynomial as product of two linear factors by splitting the middle term and then equate each factor to zero to get desired roots of given quadratic equation.
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Answer:
a) When the beam is incident on the lens from medium μ
1
.
Then
v
μ
2
−
u
μ
1
=
R
μ
2
−μ
1
or
v
μ
2
−
(−∞)
μ
1
=
R
μ
2
−μ
1
or
v
1
=
μ
2
R
μ
2
−μ
1
or v=
μ
2
−μ
1
μ
2
R
Again, for 2ns refraction,
v
μ
3
−
u
μ
2
=
R
μ
3
−μ
2
or,
v
μ
3
=−[
R
μ
3
−μ
2
−
μ
2
R
μ
2
(μ
2
−μ
1
)]
⇒−[
R
μ
3
−μ
2
−μ
2
+μ
1
]
or, v=−[
μ
3
−2μ
2
+μ
1
μ
3
R
]
So, the image will be formed at =
2μ
2
−μ
1
−μ
3
μ
3
R
b) Similarly for the beam from μ
3
medium the image is formed at
2μ
2
−μ
1
−μ
3
μ
1
R
.