Question

Find the zero of the root of the following:- 1. 2x^2+5x+7​

Answer 1

$2 {x}^{2} + 5x + 7 = 0$

Using the Quadratic Formula,

$x = \frac{ - b ± \sqrt{ {b}^{2} - 4ac } }{2a}$

$x = \frac{ - 5 ± \sqrt{ {5}^{2} - 4(2)(7) } }{2(2)} \\ x = \frac{ - 5 ± \sqrt{25 - 56} }{4} \\ x = \frac{ - 5 ± \sqrt{ - 31} }{4} \\ x = \frac{ - 5 ± i \sqrt{31} }{4} \\</p><p>x = \frac{ - 5 + i \sqrt{31} }{4} \\</p><p>x = \frac{ - 5 - i \sqrt{31} }{4}$

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Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{x=\frac{-5\pm\sqrt{31}\iota}{4}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies {2x}^{2} + 5x + 7 = 0 \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Zeroes \: of \: quadratic = ?$

• According to given question :

$\tt \circ \: {2x}^{2} + 5x + 7 = 0 \\ \\ \tt \circ \: a = 2 \\ \\ \tt \circ \: b = 5 \\ \\ \tt \circ \: c = 7 \\ \bold{As \: we \: know \: that} \\ \tt: \implies D = {b}^{2} - 4ac \\ \\ \tt: \implies D = {5}^{2} - 4 \times 7 \times 2 \\ \\ \tt: \implies D = 25 - 56 \\ \\ \tt: \implies D = - 31 \\ \\ \tt: \implies x= \frac{ - b \pm \sqrt{D} }{2a} \\ \\ \tt: \implies x = \frac{ - 5 \pm \sqrt{ - 31} }{2 \times 2 } \\ \\ \tt \circ \: \sqrt{ - 31} = \sqrt{31} \iota \\ \\ \green{\tt: \implies x = \frac{ - 5 \pm \sqrt{31} \iota }{4} }$