Math, asked by prachi7255, 10 months ago

(5) 5m² + 2m + 1 = 0

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Answered by arvishaali2004

Answer:

Step-by-step explanation:

The quadratic eqn will not have any real roots, only it posses imaginary roots.

By Quadratic formula,

In this eqn, a=5, b=2, and c=1....

So, b^2-4ac= 2^2-4*5*1

=> 4-20= -16<0

So, it is impossible to have real roots, as b^2-4ac<0........

Answered by abdul9838

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$\bf \red{hey \: user \: here \: is \: ans} \\ \\ \bf \blue{ \huge \: solution} \\ \\ \bf \green{given \: that} \\ \\ \bf \orange{5 {m}^{2} + 2m + 1 = 0 } \\ \\ \bf \pink{ here \: is \: ans} \\ \\ \bf \purple{method \: = completing \: square} \\ \\ \bf \orange{5 {m}^{2} + 2m + 1 = 0 } \\ \\ \bf \purple{ \underline{multiplying \: by \: 5 \: into \: both \: sides}} \\ \\ \bf \orange{25 {m}^{2} + 10m + 5 = 0 \times 5} \\ \\ \bf \orange{ ({5m)}^{2} + 2 \times 5m \times 1 + {1}^{2} - {1}^{2} + 5 = 0 } \\ \\ \bf \orange{( {5m + 1)}^{2} - 1 + 5 = 0 } \\ \\ \bf \orange{( {5m + 1)}^{2} + 4 = 0 } \\ \\ \bf \orange{(5m + 1)^{2} = \pm \: 4} \\ \\ \bf \orange{5m + 1 = \pm \: \sqrt{4} } \\ \\ \bf \orange{5m + 1 = \pm \: 2} \\ \\ \bf \orange{5m = 2 - 1} \\ \\ \bf \orange{m = \frac{1}{5} } \\ \\ \bf \orange{and} \\ \\ \bf \orange{5m = - 2 - 1 } \\ \\ \bf \orange{m = - \frac{3}{5} \: \: ans}$

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(5) 5m² + 2m + 1 = 0