Question

y
${y}^{2} + 6y + 63 = 0$
​

Answer 1

Question :-

Find for y ,

$\mapsto \sf {y}^{2} + 6y + 63 = 0$

Answer :-

$\to \boxed{\sf \: y = - 3 \pm \iota \: \sqrt{54} } \sf \: imaginary \: value \\ \sf \: or \: \\ \to \boxed{\sf \: y = - 3 \pm \sqrt{ - 54} } \:$

Used formula :-

Here we will use shri dharacharya principle .

If we have a Quadratic f(y) = ay² + by + c = 0

i,e.

$\to \sf \: y \: = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a}$

Solution :-

We have ,

$\to \sf \: {y}^{2} + 6y + 63 = 0 \\ \\ \sf \: here \: \\ \to \sf \: a = 1 , b = 6 \: \: and \: c = 63 </p><p> \: \\ \therefore \: \\ \\ \to \sf y = \frac{ - 6 \pm \sqrt{ {(6)}^{2} - 4 \times 1 \times 63} }{2 \times 1} \\ \\ \to \sf \: y = \frac{ - 6 \pm \sqrt{36 - 252} }{2} \\ \\ \sf \to \: y = \frac{ - 6 \pm \sqrt{ (- 216}) }{2} \\ \\ \to \sf y = \frac{ - 6 \pm \: \sqrt{ - 1} \sqrt{216} }{2} \\ \\ \: \: \: \: \: \: \boxed{ \sf \because \iota(iota) = \sqrt{ - 1} } \\ \\ \to \sf \: y = \frac{ - 6 \pm \iota \sqrt{216} }{2} \\ \\ \to \sf y = \frac{ - 6 \pm \: 2 \iota \sqrt{54} }{2} \\ \\ \to \boxed{\sf \: y = - 3 \pm \iota \: \sqrt{54} } \sf \: imaginary \: value \\ \sf \: or \: \\ \\ \to \boxed{\sf \: y = - 3 \pm \sqrt{ - 54} }$