Question

Ifa and Bare the zeros of the polynomial f(x) = x2 + px +q, form a polynomial whose
zeros are (a + b)2 and (a - b).
are the terms of the quadratic polynomial f(x) = x2 - 2x + 3, find a
0​

Answer 1

Step-by-step explanation:

${\bf{Need~To~Prove~:}~\rm{L.H.S~=~\dfrac{\bigg(\cos\theta \cancel{+} \sin\theta\bigg)\bigg(\cos^2 \theta + \cos\theta\sin\theta + \sin^2 \theta\bigg)}{\bigg(\cos\cancel{\theta} + \sin\theta\bigg)}~\bf{+}~\dfrac{\bigg(\cos\theta \cancel{-} \sin\theta\bigg)\bigg(\cos^2 \theta - \cos\theta\sin\theta + \sin^2 \theta\bigg)}{\bigg(\cos\theta \cancel{-} \sin\theta\bigg)}}}Need To Prove : L.H.S = (cosθ+sinθ)(cosθ+sinθ)(cos2θ+cosθsinθ+sin2θ) + (cosθ−sinθ)(cosθ−sinθ)(cos2θ−cosθsinθ+sin2θ)</p><p></p><p>~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~{\sf:\implies{\cos^2 \theta + \cos\cancel{\theta} \sin\theta + \sin^2 \theta + \cos^2 \theta - \cos\cancel{\theta} \sin\theta + \sin^2 \theta}}                                      :⟹cos2θ+cosθsinθ+sin2θ+cos2θ−cosθsinθ+sin2θ</p><p></p><p>~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~{\sf:\implies{2 \cos^2 \theta + 2 \sin^2 \theta}}                                      :⟹2cos2θ+2sin2θ</p><p></p><p>~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~{\sf:\implies{2 \bigg(\cos^2 \theta + 2 \sin^2 \theta\bigg)}}                                      :⟹2(cos2θ+2sin2θ)</p><p></p><p>~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~:\implies{\underline{\boxed{\frak{\pink{2~=~R.H.S}}}}}                                      :⟹2 = R.H.S</p><p></p><p>$