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If the zeros of the polynomial …

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If the zeros of the polynomial f(x)=2xcube +15xsquare 37x-30 are in A.P.,Find them.

  • 1 answers

Sia ? 4 years, 10 months ago

Let, {tex}α{/tex} = a - d, {tex}β{/tex} = a and  {tex}\gamma {/tex} = a + d be the zeroes of the polynomial.
f(x) = 2x3 - 15x2 + 37x - 30
{tex}\alpha + \beta + \gamma = - \left( {\frac{{ - 15}}{2}} \right) = \frac{{15}}{2}{/tex} ....... (i)
{tex}\alpha \beta \gamma = - \left( {\frac{{ - 30}}{2}} \right) = 15{/tex} ......... (ii)
From (i)
a - d + a + a + d = {tex}\frac{{15}}{2}{/tex} 
So, 3a = {tex}\frac{{15}}{2}{/tex}
a = {tex}\frac{{5}}{2}{/tex}
and From (ii)
a(a - d)(a + d) = 15
So, a(a2 - d2) = 15
{tex}⇒ \frac{{5}}{2}{/tex} {tex}\left[\left(\frac52\right)^2\;-d^2\right]{/tex} = 15
{tex}⇒ \frac{25}4\;-d^2{/tex}= 6
{tex}⇒ \;d^2\;=\;\frac{25}4-6\;{/tex}
{tex}⇒ {d^2} = \frac{1}{4}{/tex}
{tex}⇒ d = \frac{1}{2}{/tex}
Therefore, {tex}\alpha = \frac{5}{2} - \frac{1}{2} = \frac{4}{2} = 2{/tex}
{tex}\beta = \frac{5}{2}{/tex}
{tex}\gamma = \frac{5}{2} + \frac{1}{2} = 3{/tex}.

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