Question

S
(0) Solve the canation by quadratic formula
(i) x² +50+5=0 (i) x2 +6x +6=0
MARCH​

Answer 1

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Q1) Solve given quadratic equation by formula method ::

$\: \bf{(1) \: \: {x}^{2} + 50x + 5 = 0}$

$\: \quad \implies \underline{ \pink \odot \: \displaystyle \sf{by \: determinant \: form}}$

$\: \quad \implies \displaystyle \sf{ {b}^{2} - 4ac}$

$\: \: \quad \implies \displaystyle \sf{ {( - 50)}^{2} - 4(1)(5)}$

$\: \quad \implies \displaystyle \sf{2500 - 20}$

$\: \: \quad \implies \displaystyle \sf{2480}$

$\: \: \quad \implies \underline{ \ast\displaystyle \sf{ \: by \: formula \: method}}$

$\: \: \quad \implies \displaystyle \sf{x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} }$

$\: \quad \implies \displaystyle \sf{x = \frac{ - ( - 50) \pm \sqrt{2480} }{2 \times 1} }$

$\: \: \quad \implies \displaystyle \sf \: x = \frac{50 \pm \sqrt{2480} }{2}$

$\: \: \quad \implies \displaystyle \sf \: x = \frac{50 \pm2 \sqrt{620} }{2}$

$\: \: \quad \implies \displaystyle \sf \: x = \frac{2(25 \pm \sqrt{620} ) }{2}$

$\: \: \quad \implies \displaystyle \sf \: x = \cancel \frac{2}{2} \bigg( \frac{25 \pm \sqrt{620} }{1} \bigg)$

$\: \: \quad \implies \displaystyle \sf \: x = 25 \pm \sqrt{620}$

$\: \: \boxed{ \boxed{ \underline{ \bf{ \color{red} \: the \: roots \: are \: 25 + 2 \sqrt{155} \: and \: 25 - 2 \sqrt{155} }}}}$

$\pink{▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬}$

$\: \bf{(2) \: \: {x}^{2} + 6x + 6 = 0}$

$\: \\ \quad \implies \displaystyle \sf{ {x}^{2} + 6x + 6 = 0 \: ....given \: quadratic \: equation}$

$\: \: \quad \implies \underline{ \displaystyle \sf{ \color{orange} \diamondsuit \: by \: determinant \: form : }}$

$\: \: \quad \implies \displaystyle \sf \: \delta \: d= {b}^{2} - 4ac$

$\: \: \quad \implies \displaystyle \sf \: {( - 6)}^{2} - 4( 1)( - 6)$

$\: \: \quad \implies \displaystyle \sf \: 36 + 24$

$\: \: \quad \implies \displaystyle \sf \: 60$

$\: \: \\ \: \quad \implies \displaystyle \sf{ \pink \clubsuit \: by \: using \: formula \: method}$

$\: \: \quad \implies \displaystyle \sf \: x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a}$

$\: \: \quad \implies \displaystyle \sf \: x = \frac{ -6 \pm \sqrt{60} }{2 \times 1}$

$\: \: \quad \implies \displaystyle \sf \: x = \frac{ - 6 \pm \sqrt{60} }{2}$

$\: \: \quad \implies \displaystyle \sf \: x = \frac{ - 6 \pm \sqrt{4 \times 15} }{2}$

$\: \: \quad \implies \displaystyle \sf{x = \frac{ - 6 \pm2 \sqrt{15} }{2} }$

$\: \: \quad \implies \displaystyle \sf \: x = \frac{ 2( - 3 \pm \sqrt{15}) }{2}$

$\: \: \quad \implies \displaystyle \sf x = \cancel \frac{2}{2} \frac{ - 3 \pm \sqrt{15} }{1}$

$\: \: \quad \implies \displaystyle \sf{x = - 3 \pm \sqrt{15}}$

$\: \boxed{ \boxed{ \underline{ \underline{ \color{brown} \bf{the \: roots \: are \: - 3 + \sqrt{15} \: and \: x = - 3 - \sqrt{15} }}}}}$

$\color{lime}{▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬}$