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Sia ? 6 years, 6 months ago
{tex}(a + b)^2x^2 + 8(a^2 - b^2)x + 16(a - b)^2 = 0{/tex}
{tex}A = (a + b)^2,\ B = 8(a^2 - b^2),\ C = 16(a - b)^2 {/tex}
D = B2 - 4AC
= [8(a2 - b2)]2 - 4 {tex}\times{/tex} (a + b)216(a - b)2
= 64(a2 - b2)2 - 64(a2 - b2)2 = 0
{tex}\therefore{/tex}x = {tex}\frac{-B}{2A}{/tex} = {tex}\frac { - 8 \left( a ^ { 2 } - b ^ { 2 } \right) } { 2 ( a + b ) ^ { 2 } }{/tex} = {tex}\frac { - 4 ( a - b ) } { a + b }{/tex}
Posted by Prem Bisht 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Let us assume, to the contrary, that √p is rational.
So, we can find co-prime integers a and b(b ≠ 0)
{tex}\begin{array}{l}\sqrt p=\frac ab\\\end{array}{/tex}
{tex}a = b _ { \sqrt { P } }{/tex}
on squaring both sides we get
a2 = pb2 ...... (1)
so a2 is divisible by p
hence a is divisible by p ....... (2)
So, we can write a = pc for some integer c.
Squaring both the sides we get
a2 = p2 c2 ....
⇒ pb2 = p2 c2 ....[From (1)]
⇒ b2 = pc2
⇒ b2 is divisible by p
⇒ b is divisible by p ....... (3)
From (2) and (3) we conclude that p divides both a and b.
∴ a and b have at least p as a common factor.
But this contradicts the fact that a and b are co-prime. (As per our assumption)
This contradiction arises because we have assumed that √p is rational.
∴ √p is irrational.
Posted by Abhiraj Jain 7 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let the required number be (a - 3d), (a - d), (a + d) and (a + 3d)
Sum of these numbers = (a - 3d) + (a - d)+ (a + d) + (a + 3d)
According to the question, sum of the numbers=32
{tex}\therefore{/tex}4a = 32 {tex}\Rightarrow{/tex} a = 8
Sum of the squares of these numbers
=(a-3d)2+(a-d)2+(a+d)2+(a+3d)2=4(a2+5d{tex}^2{/tex})
Now, sum of the squares of numbers=336
{tex}\therefore{/tex}4(a2+5d2)=336
{tex}\Rightarrow{/tex}a2+5d2=84 [{tex}\because {/tex}a=8]
{tex}\Rightarrow{/tex}5d2= 84-64
{tex}\Rightarrow{/tex}5d2=20
{tex}\Rightarrow{/tex}d2=4
{tex}\Rightarrow{/tex}d={tex} \pm {/tex}2
Hence, the required numbers (2, 6, 10, 14).
Posted by Harshita Sharma 7 years, 6 months ago
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