Question

18-3y/4 = 11+ y please help me with this sum .​

Answer 1

The roots are $\alpha =\frac{9+i\sqrt{51} }{3}$ and $\beta =\frac{9-i\sqrt{51} }{3}$

Explanation:

Given:

$\frac{18-3y}{4}=\frac{11}{y}$

To find:

The value of y

Formula:

$\alpha =\frac{-b+\sqrt{b^{2}-4ac } }{2a}$

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

Solution:

==> $\frac{18-3y}{4}=\frac{11}{y}$

==> $18-3y=\frac{4\times11}{y}$

==> $y(18-3y)=4\times11$

==> $18y-3y^{2}=44$

==> $18y-3y^{2}-44=0$

==> Multiply by - on both sides

==> $-18y+3y^{2}+44=0$

==> $3y^{2}-18y+44=0$

==> a = coeffcient of y²

==> b = coeffcient of y

==> c = Constant

==> a = 3

==> b = -18

==> c = 44

==> Substitute the values in the formula

==> $\alpha =\frac{-(-18)+\sqrt{(-18)^{2}-4(3)(44) } }{2(3)}$

==>  $\alpha =\frac{18+\sqrt{(324-12(44) } }{6}$

==> $\alpha =\frac{18+\sqrt{(324-528 } }{6}$

==> $\alpha =\frac{18+\sqrt{-204 } }{6}$

==> we know that, i² = -1

==> $\alpha =\frac{18+\sqrt{-1\times 204 } }{6}$

==> $\alpha =\frac{18+\sqrt{-1\times 2\times2\times3\times17} }{6}$

==> $\alpha =\frac{18+\sqrt{i^{2} \times 2\times2\times3\times17} }{6}$

==> $\alpha =\frac{18+2i\sqrt{51} }{6}$

==> $\alpha =\frac{2(9+i\sqrt{51)} }{6}$

==> $\alpha =\frac{9+i\sqrt{51} }{3}$

==> $\beta =\frac{-(-18)-\sqrt{(-18)^{2}-4(3)(44) } }{2(3)}$

==>  $\beta =\frac{18-\sqrt{(324-12(44) } }{6}$

==> $\beta =\frac{18-\sqrt{(324-528 } }{6}$

==> $\beta =\frac{18-\sqrt{-204 } }{6}$

==> we know that, i² = -1

==> $\beta =\frac{18-\sqrt{-1\times 204 } }{6}$

==> $\beta =\frac{18-\sqrt{-1\times 2\times2\times3\times17} }{6}$

==> $\beta =\frac{18-\sqrt{i^{2} \times 2\times2\times3\times17} }{6}$

==> $\beta =\frac{18-2i\sqrt{51} }{6}$

==> $\beta =\frac{2(9-i\sqrt{51)} }{6}$

==> $\beta =\frac{9-i\sqrt{51} }{3}$

The roots are $\alpha =\frac{9+i\sqrt{51} }{3}$ and $\beta =\frac{9-i\sqrt{51} }{3}$