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Sia ? 6 years, 5 months ago
Given that -5 is the root of {tex}2 x^{2}+p x-15=0{/tex}
Put x = -5 in {tex}2 x^{2}+p x-15=0{/tex}
{tex}\Rightarrow{/tex} {tex}2(-5)^{2}+p(-5)-15=0{/tex}
{tex}\Rightarrow{/tex} {tex}50-5 p-15=0{/tex}
{tex}\Rightarrow{/tex} {tex}35 - 5p = 0 {/tex}
{tex}\Rightarrow{/tex} {tex}5p = 35 {/tex}
{tex}\therefore{/tex} {tex}p = 7{/tex}
Hence the quadratic equation p {tex}(x^2 + x) + k = 0{/tex} becomes, {tex}7\left(x^{2}+x\right)+k=0{/tex}
{tex}7 x^{2}+7 x+k=0{/tex}
Here {tex}a = 7,\ b = 7\ and\ c = k{/tex} Given that this quadratic equation has equal roots
{tex}\therefore{/tex} {tex}b^{2}-4 a c=0{/tex}
{tex}\Rightarrow{/tex} {tex}7^{2}-4(7)(\mathrm{k})=0{/tex}
{tex}\Rightarrow{/tex} {tex}49 - 28k = 0 {/tex}
{tex}\Rightarrow{/tex} {tex}49 = 28k {/tex}
{tex}\therefore{/tex} k = {tex}\frac{49} {28}{/tex} = {tex}\frac{7} {4}{/tex}
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Sia ? 6 years, 4 months ago
A, B, C, are interior angles of a {tex}\Delta {/tex}
{tex}\because A + B + C = 180 ^ { 0 }{/tex}
{tex}\Rightarrow B + C = 180 ^ { 0 } - A \Rightarrow \frac { B + C } { 2 } = 90 ^ { 0 } - \frac { A } { 2 }{/tex}
{tex}\Rightarrow \sin \frac { \mathrm { B } + \mathrm { C } } { 2 } = \sin \left( 90 ^ { \circ } - \frac { \mathrm { A } } { 2 } \right) \left[ \because \sin \left( 90 ^ { 0 } - \theta \right) = \cos \theta \right]{/tex}
{tex}\Rightarrow \sin \frac { \mathrm { B } + \mathrm { C } } { 2 } = \cos \frac { \mathrm { A } } { 2 } \text { proved }{/tex}
LHS = RHS
Posted by Vishal Kumar Patel 7 years, 4 months ago
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