Question

find the nature of root of quadratic equation
$4x {}^{2} + 6x + 1 = 0$
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Answer 1

Answer:

$4 {x}^{2} + 6x + 1$

Step-by-step explanation:

$a = 4 \\ b = 6 \\ c = 1 \\ quadratic \: formula = ( - b + - \sqrt{ {b - 4ac}^{2} } ) \div 2a$

$= ( - 6 + - \sqrt{ {6}^{2} - 4(4)(1) } ) \div 2(4) \\ = (- 6 + - \sqrt{36 - 16} ) \div 8 \\ x1 = ( - 6 + \sqrt{20} ) \div 8 \\ x1 = - 0.19 \\ x2 = ( - 6 - \sqrt{20} ) \div 8 \\ x2 = - 1.31$

So, the roots are real but not perfect square