InterviewSolution
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InterviewSolution
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Let `a ,b ,c`be real numbers with `a!=0a n dl e talpha,beta`be the roots of the equation `a x^2+b x+c=0.`Express the roots of `a^3x^2+a b c x+c^3=0`in terms of `alpha,betadot` |
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Answer» Correct Answer - `x=alpha^(2)beta,alphabeta^(2)` Since, `ax^(2)+bx+c=0` has roots `alpha and beta.` `impliesalpha+beta=-b//a` and ` alphabeta=c//a` `Now, a^(3)x^(2)+abcx+c^(3)=0" "...(i)` On dividing the equation by `c^(2),` we get `(a^(3))/(c^(2))x^(2)(abcx)/(c^(2))+(c^(3))/(c^(2))=0` `impliesa((ax)/(c))^(2)+b((ax)/(c))+c=0` `impliesx=c/aalpha,c/abeta` are the roots `implies x=alphabeta, alpha beta beta` are the roots `impliesx=alpha^(2)beta,alphabeta^(2)` are teh roots Divide the Eq. (i) by `a^(3),` we get `x^(2)+b/c.c/ax+((c)/(a))^(3)=0` `impliesx^(2)-(alpha+beta)(alphabeta)x+(alphabeta)^(3)=0` `impliesx^(2)-alpha^(2)betax-alpha beta ^(2)x+(alphabeta)^(2)=0` `impliesx(x-alpha^(2)beta)-alphabeta^(2)(x-alphabeta^(2))=0` `implies (x-alpha^(2)beta)(x-alphabeta^(2))=0` `impliesx=alpha^(2)beta,alpha beta^(2)` which is the required answer. Alternate Solution Since, `a^(3)x^(2)+abcx+c^(3)=0` `impliesx=(-abcpmsqrt((abc)^(2)-4.a^(3).c^(3)))/(2a^(3))` `impliesx=(-(b//a)(c//a)pmsqrt((b//c)^(2)(c//a)^(2)-4(c//a)^(3)))/(2)` `impliesx=((alpha+beta)(alphabeta)pmsqrt((alpha+beta)^(2)(alphabeta)^(2)-4(alphabeta)^(3)))/(2)` `impliesx=((alpha+beta)(alphabeta)pmalphabetasqrt((alpha+beta)^(2)-4alphabeta))/(2)` `impliesx=alphabeta[((alpha+beta)pmsqrt((alpha-beta)^(2)))/(2)]` `impliesx=alphabeta[((alpha+beta)pmsqrt((alpha-beta)))/(2)]` `impliesx=alphabeta[(alpha+beta+alpha-beta)/(2),(alpha+beta-alpha+beta)/(2)]` `impliesx=alpha beta[(2alpha)/(2),(2 beta)/(2)]` `impliesx=alpha^(2)beta, alpha beta^(2)` which is the required answer. |
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