Solve x (2x^2/x-5)+(10x/x-5)-24=0
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Sia ? 6 years, 5 months ago
We have given,
{tex}\left( \frac { 2 x } { x - 5 } \right) ^ { 2 } + 5 \left( \frac { 2 x } { x - 5 } \right) - 24 = 0{/tex}
Let {tex}\frac { 2 x } { ( x - 5 ) } \text { be } y{/tex}
{tex}\therefore \quad y ^ { 2 } + 5 y - 24 = 0{/tex}
Now factorise,
{tex}y^2+8y-3y-24=0{/tex}
{tex}y(y+8)-3(y+8)=0{/tex}
{tex}(y+8)(y-3)=0{/tex}
{tex}y=3,-8{/tex}
Putting y=3
{tex}\frac { 2 x } { x - 5 } = 3{/tex}
2x = 3x - 15
x = 15
Putting y = -8
{tex}\frac { 2 x } { x - 5 } = - 8{/tex}
2x = -8x + 40
10x = 40
x = 4
Hence, x is 15 , 4
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