Answer:
Given, \[5\] is a root of \[2{{x}^{2}}+px-15=0\] then, \[f(-5)=2{{(-5)}^{2}}+p(-5)-15=0\] \[50-5p-15=0\] \[35-5p=0\] \[5p=35\] \[p=7\] Now, putting the value of p, in, \[p({{x}^{2}}+x)+k=0\] we get \[7{{x}^{2}}+7x+k=0\] Now, \[D={{b}^{2}}-4ac=0\] (\[\because \] has the equal roots) then, \[49-28k=0\] \[28k=49\] \[k=\frac{49}{28}=\frac{7}{4}\]
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