EXAMPLE 11
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ABCD is a rhombous in which angle A = angle B - 30 then ,find angle C
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a, b, c and p are rational. p is not a perfect cube.
so p¹/³ and p²/³ are not rational numbers. Then p is not 1 or 0.
p²/³ = p¹/³ * p¹/³ and so they are not equal.
LHS = a + b p¹/³ + c p²/³ = 0 --- (1) \begin{lgathered}a+b p^\frac{1}{3}+ c p^\frac{2}{3}=0. .. ----(1)\\\end{lgathered}a+bp31+cp32=0...−−−−(1)Given p is not a perfect cube. p is not 0 or 1. Also p^\frac{1}{3} \: p^\frac{2}{3}p31p32 are irrational. Multiply (1) by p^1/3 to get: \begin{lgathered}cp + a p^\frac{1}{3} + b p^\frac{2}{3} = 0. \: ... ... (2)\\(1) \: & \: (2) : \frac{a}{cp} = \frac{b}{a} = \frac{c}{b}\\\\a^2=cpb \: and \: b^2=ac. ... ...(3)\\c = \frac{b^2}{a}.\end{lgathered}cp+ap31+bp32=0.......(2)(1)a2=cpbandb2=ac.......(3)c=ab2.(2):cpa=ab=bcSubstitute the value of c in (1) to get: \begin{lgathered}a^2 + a b p^\frac{1}{3} + b^2 p^\frac{2}{3} = 0. .. Quadratic \\\\Discriminant = -3 a^2 b^2 = negative.\end{lgathered}a2+abp31+b2p32=0...QuadraticDiscriminant=−3a2b2=negative.So p^1/3 is imaginary. It is a contradiction as p is a rational number. Given quadratic isn't valid. So a = b = c = 0.
so p¹/³ and p²/³ are not rational numbers. Then p is not 1 or 0.
p²/³ = p¹/³ * p¹/³ and so they are not equal.
LHS = a + b p¹/³ + c p²/³ = 0 --- (1) \begin{lgathered}a+b p^\frac{1}{3}+ c p^\frac{2}{3}=0. .. ----(1)\\\end{lgathered}a+bp31+cp32=0...−−−−(1)Given p is not a perfect cube. p is not 0 or 1. Also p^\frac{1}{3} \: p^\frac{2}{3}p31p32 are irrational. Multiply (1) by p^1/3 to get: \begin{lgathered}cp + a p^\frac{1}{3} + b p^\frac{2}{3} = 0. \: ... ... (2)\\(1) \: & \: (2) : \frac{a}{cp} = \frac{b}{a} = \frac{c}{b}\\\\a^2=cpb \: and \: b^2=ac. ... ...(3)\\c = \frac{b^2}{a}.\end{lgathered}cp+ap31+bp32=0.......(2)(1)a2=cpbandb2=ac.......(3)c=ab2.(2):cpa=ab=bcSubstitute the value of c in (1) to get: \begin{lgathered}a^2 + a b p^\frac{1}{3} + b^2 p^\frac{2}{3} = 0. .. Quadratic \\\\Discriminant = -3 a^2 b^2 = negative.\end{lgathered}a2+abp31+b2p32=0...QuadraticDiscriminant=−3a2b2=negative.So p^1/3 is imaginary. It is a contradiction as p is a rational number. Given quadratic isn't valid. So a = b = c = 0.
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