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Gaurav Seth 6 years, 10 months ago
Given
px2-14x+8=0
let a and b be the roots of the equation
b=6a
sum of roots = -b/a
product of roots = c/a
here a=p , b=-14 , c=8
a+b=14/p
ab=8/p
a+6a=14/p
7a=14/p
a=2/p
a*6a=8/p
6a2=8/p
3a2=4/p
3*(2/p)2=4/p
3*4/p2=4/p
p=3
Posted by Jitesh Baviskar 6 years, 10 months ago
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Gaurav Seth 6 years, 10 months ago
Alternate Segment:
If a chord is drawn through the point of contact of a tangent to a circle, then the angles which this chord makes with the given tangent are equal respectively to the angles formed in the corresponding alternate segments.
Given : 
is the tangent to the circle with centre O. AB is the chord drawn at the point of contact A.
C and D are two points on the circumference such that they lie on either side of the chord.
To Prove: (i) 
and (ii) 
Construction: Draw the diameter AOE and draw EB.
Proof: In the figure, 
(
radius is perpendicular to the tangent at the point of contact)
(1)
In DEAB
(Angle in a semi-circle)
(2)
From (1) and (2)
But, 
(Angles in the same segment)
Now, consider cyclic quadrilateral ACBD.
(Opposite angles)
.(4)
And 
(Linear pair)
But
(5)
From (4) and (5)
Hence proved.
Posted by Aashrity .... 6 years, 10 months ago
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Ram Kushwah 6 years, 10 months ago
X,Y,Z are in AP
then y-x=z-y
2y=x+z
y=x+z/2
According to question
(x+y-z)(-x+y+z)
x-z+x+z (-x+x+z +z)
2 2
(2x-2z+x+z) (x+z+2z-2x)
2 2
1/4 (3x-z)(3z-x)
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Purvanshi Yadav 6 years, 10 months ago
0Thank You