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Yogita Ingle 5 years, 7 months ago
Let us assume that √3 is a rational number.
then, as we know a rational number should be in the form of p/q
where p and q are co- prime number.
So,
√3 = p/q { where p and q are co- prime}
√3q = p
Now, by squaring both the side
we get,
(√3q)² = p²
3q² = p² ........ ( i )
So,
if 3 is the factor of p²
then, 3 is also a factor of p ..... ( ii )
=> Let p = 3m { where m is any integer }
squaring both sides
p² = (3m)²
p² = 9m²
putting the value of p² in equation ( i )
3q² = p²
3q² = 9m²
q² = 3m²
So,
if 3 is factor of q²
then, 3 is also factor of q
Since
3 is factor of p & q both
So, our assumption that p & q are co- prime is wrong
hence,. √3 is an irrational number
Posted by Santosh Upadhyay 5 years, 7 months ago
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Yogita Ingle 5 years, 7 months ago
We can do it by using Pythagoras Theorem.
We can write √10 = √(9 + 1)
=> √10 = √(32 + 1)
Construction
1. Take a line segment AO = 3 unit on the x-axis. (consider 1 unit = 2cm)
2. Draw a perpendicular on O and draw a line OC = 1 unit
3. Now join AC with √10.
4. Take A as center and AC as radius, draw an arc which cuts the x-axis at point E.
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Yogita Ingle 5 years, 7 months ago
(x+4)(x+10)
= x(x + 10) + 4(x + 10)
= x2 + 10x + 4x + 40
= x2 + 14x +40
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Yogita Ingle 5 years, 7 months ago
f(x)=2x³-3x+7a
x = 2 is a root of the polynomial
2x3 - 3x + 7a =0
2(2)3 - 3(2) + 7a = 0
2(8) - 6 + 7a = 0
16- 6 + 7a = 0
10 + 7a = 0
7a = -10
a = -10/7
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