Find the value of p if the difference of square of the zeroes of polyn

1.	Find the value of p if the difference of square of the zeroes of polynomial x2 +px+45 is 144
Answer» If the squared difference of the zeroes of the quadratic polynomial f(x) = x2 + px + 45 is equal to 144, then ,we have to find the value of p.Let {tex}\\alpha{/tex}\xa0and {tex}\\beta{/tex}\xa0be the zeroes of the given quadratic polynomial.{tex}\\therefore{/tex}\xa0{tex}\\alpha{/tex} + {tex}\\beta{/tex} = - p and {tex}\\alpha\\beta{/tex}= 45 ...(i)Given, ({tex}\\alpha{/tex} - {tex}\\beta{/tex})2 = 144or, ({tex}\\alpha{/tex} + {tex}\\beta{/tex})2- 4{tex}\\alpha{/tex}{tex}\\beta{/tex}\xa0= 144or, (-p)2\xa0- 4 {tex}\\times{/tex}\xa045 = 144 [Using (i)]p2\xa0- 180 = 144p2 = 144 + 180 = 324{tex}\\therefore{/tex}\xa0p = ± {tex}\\sqrt{324}{/tex}= ± 18Hence,\xa0the value of p is ± 18.

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