Question

is the roots of 32 x cube minus 48 x square + 22 x minus 3 equal to zero are in AP then the middle root is​

Answer 1

$\textbf{Given:}$

$32\,x^3-48\,x^2+22\,x-3=0$

$\text{To find:}$

$\text{The middle root of the given equation}$

$\textbf{Solution:}$

$\text{Since the roots of 32\,x^3-48\,x^2+22\,x-3=0 are in A.P,}$

$\text{let the roots be a-d, a, a+d}$

$\textbf{Sum of the roots= \bf\dfrac{48}{32} }$

$\implies\,(a-d)+a+(a+d)=\dfrac{3}{2}$

$\implies\,3a=\dfrac{3}{2}$

$\implies\,a=\dfrac{1}{2}$

$\textbf{Product of the roots= \bf\dfrac{3}{32} }$

$\implies\,(a-d)a(a+d)=\dfrac{3}{32}$

$\implies\,a(a^2-d^2)=\dfrac{3}{32}$

$\implies\,\dfrac{1}{2}(\dfrac{1}{4}-d^2)=\dfrac{3}{32}$

$\implies\,\dfrac{1}{4}-d^2=\dfrac{3}{16}$

$\implies\,\dfrac{1}{4}-\dfrac{3}{16}=d^2$

$\implies\,d^2=\dfrac{1}{16}$

$\implies\,d=\pm\dfrac{1}{4}$

$\therefore\textbf{The required roots are \dfrac{1}{4},\dfrac{1}{2},\dfrac{3}{4} or \dfrac{3}{4},\dfrac{1}{2},\dfrac{1}{4} }$

$\implies\textbf{The middle root is \dfrac{1}{2} }$

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If alpha,beta are the roots of equation x^2-5x+6=0 and alpha > beta then the equation with the roots (alpha+beta)and (alpha-beta) is

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The equation whose roots are smaller by 1 than those of 2x2 – 5x + 6 = 0 is 

a) 2x^2– 9x + 13 = 0

b) 2x^2– x + 3 = 0 

c) 2x^2+ 9x + 13 = 0

d) 2x^2 + x + 3 = 0​

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