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Sia ? 6 years, 5 months ago
One zero {tex}= -5{/tex}
product of zeroes {tex}= 0{/tex}
{tex}\therefore{/tex} Other zero = {tex}\frac { 0 } { - 5 }{/tex} {tex}= 0{/tex}
Sum of zeroes {tex}= -5 + 0 = -5{/tex}
Polynomial {tex}p(x) = x^2 - (S)x + P{/tex}
= {tex}x^2 + 5x{/tex}
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- In DAPB and DAQB
{tex}\angle{/tex}BAP = {tex}\angle{/tex}BAQ ...[As l bisects {tex}\angle{/tex}A]
{tex}\angle{/tex}BPA = {tex}\angle{/tex}BQA ...[Each 90°]
AB = AB ...[Common]
{tex}\therefore{/tex} DAPB {tex}\cong{/tex} DAQB proved ...[AAS property] ...(1) - DAPB {tex}\cong{/tex} DAQB ...[From (1)]
{tex}\therefore{/tex} BP = BQ ...[c.p.c.t.]
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Sia ? 6 years, 5 months ago
Let us assume that <m:omath><m:r>√8</m:r></m:omath> is rational.
That is, we can find integers a and b (<m:omath><m:r>≠0)</m:r></m:omath> such that a and b are co-prime
{tex}\style{font-family:Arial}{\begin{array}{l}\sqrt8=\frac ab\\b\sqrt8=a\\on\;squaring\;both\;sides\;we\;get\\8b^2=a^2\end{array}}{/tex}
Therefore, a2 is divisible by 8,
it follows that a is also divisible by 8.
So, we can write a = 8c for some integer c.
Substituting for a, we get 8b2 = 64c2, that is, b2 = 8c2
This means that b2 is divisible by 8, and so b is also divisible by 8
Therefore, a and b have at least 8 as a common factor.
But this contradicts the fact that a and b are co-prime.
This contradiction has arisen because of our incorrect assumption that <m:omath><m:r>√8</m:r></m:omath> is rational.
So, we conclude that <m:omath><m:r>√8</m:r></m:omath> is irrational.
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Ashnoor Kaur 6 years, 5 months ago
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