If the squared difference of quadratic …

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Sia ? 6 years, 6 months ago
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} and {tex}\beta{/tex} be the zeroes of the given quadratic polynomial.
{tex}\therefore{/tex} {tex}\alpha{/tex} + {tex}\beta{/tex} = - p and {tex}\alpha\beta{/tex}= 45 ...(i)
Given, ({tex}\alpha{/tex} - {tex}\beta{/tex})2 = 144
or, ({tex}\alpha{/tex} + {tex}\beta{/tex})2- 4{tex}\alpha{/tex} {tex}\beta{/tex} = 144
or, (-p)2 - 4 {tex}\times{/tex} 45 = 144 [Using (i)]
p2 - 180 = 144
p2 = 144 + 180 = 324
{tex}\therefore{/tex} p = ± {tex}\sqrt{324}{/tex}= ± 18
Hence, the value of p is ± 18.
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