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Sia ? 6 years, 6 months ago
We have,
4x2 + 3x + 5 = 0
Divide whole equation by 4
{tex} \Rightarrow \quad x ^ { 2 } + \frac { 3 } { 4 } x + \frac { 5 } { 4 } = 0{/tex}
{tex}\Rightarrow \quad x ^ { 2 } + 2 \left( \frac { 3 } { 8 } x \right) = - \frac { 5 } { 4 }{/tex}
Add ( half of coefficient of x )2 both sides
{tex}\Rightarrow \quad x ^ { 2 } + 2 \left( \frac { 3 } { 8 } \right) x + \left( \frac { 3 } { 8 } \right) ^ { 2 } = \left( \frac { 3 } { 8 } \right) ^ { 2 } - \frac { 5 } { 4 }{/tex}
{tex}\Rightarrow \quad \left( x + \frac { 3 } { 8 } \right) ^ { 2 } = - \frac { 71 } { 64 }{/tex}
{tex}\left( x + \frac { 3 } { 8 } \right) ^ { 2 }{/tex} cannot be negative for any real value of x. Hence, the given equation has no real roots.
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