Question

Practice Set 2.3
ic equations by completing the
(2) x² + 2x - 5=0 de
(5) 2y2 + 9y + 10 = 0​

Answer 1

$\textit{\large{\underline{\underline{EXPLANATION:}}}}$

$\textbf{\underline{Que 2,,}}</p><p>$

${x}^{2} + 2x - 5 = 0 \\ = > {(x)}^{2} + 2 \times x \times 1 + {1}^{2} - 5 = 1 {}^{2} \\ = > (x + 1) {}^{2} = 1 + 5 \\ = > x + 1 = \pm \sqrt{6} \\ = > x = \pm( \sqrt{6} - 1) \\ = > x = \sqrt{6} - 1 \: or \: 1 - \sqrt{6}$

$\textbf{\underline{Que 5,,}}</p><p>$

$2 {y}^{2} + 9y + 10 = 0 \\ = > ( \sqrt{2} y) {}^{2} + 2 \times \sqrt{2} y \times \frac{9}{2 \sqrt{2} } + {( \frac{9}{2 \sqrt{2} } })^{2} = - 10 + ( \frac{9}{2 \sqrt{2} }) {}^{2} \\ = > ( \sqrt{2} y + \frac{9}{2 \sqrt{2} } ) {}^{2} = - 10 + \frac{81}{8} \\ = > {( \sqrt{2}y + \frac{9}{2 \sqrt{2} } )}^{2} = \frac{1}{8} \\ = > \sqrt{2} y + \frac{9}{2 \sqrt{2} } = \pm\sqrt{ \frac{1}{8} } \\ = > \sqrt{2} y = \frac{1}{2 \sqrt{2} } \pm \frac{9}{2 \sqrt{2} } \\ = > \sqrt{2} y = \frac{ \pm 8}{2 \sqrt{2} } \\ = > y = \frac{8}{4} \: or \: \frac{ - 8}{4} \\ = > y = 2 \: or - 2$

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