Question

X^2-6x+11 solove by quadratic formula​

Answer 1

The roots of x²-6x+11 are  3+i√2 and 3-i√2

Explanation:

Given:

x²-6x+11

Formula:

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

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

==> The quadratic equation x²-6x+11

==> x²-6x+11

==> a = coefficient of x²

==> b = coeeficient of x

==> c = constant

==> a= 1

==> b= -6

==> c=11

==> Apply these values in the formula

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

==> $\alpha =\frac{-(-6)+\sqrt{(-6)^{2}-4(1)(11) } }{2(1)}$

==>$\alpha =\frac{6+\sqrt{36-4(11) } }{2(1)}$

==>$\alpha =\frac{6+\sqrt{36-44 } }{2}$

==>$\alpha =\frac{6+\sqrt{-8 } }{2}$

==>$\alpha =\frac{6+\sqrt{-1\times2\times2\times2\times2 } }{2}$

==>$\alpha =\frac{6+2\sqrt{-1\times2 } }{2}$

==> We know that, i²=-1

==> $\alpha =\frac{6+2\sqrt{i^{2} \times2 } }{2}$

==> $\alpha =\frac{2(3+\sqrt{i^{2} \times2 }) }{2}$

==> α= 3+i√2

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

==> $\beta=\frac{-(-6)-\sqrt{(-6)^{2}-4(1)(11) } }{2(1)}$

==>$\beta=\frac{6-\sqrt{36-4(11) } }{2(1)}$

==>$\beta =\frac{6-\sqrt{36-44 } }{2}$

==>$\beta =\frac{6-\sqrt{-8 } }{2}$

==>$\beta =\frac{6-\sqrt{-1\times2\times2\times2\times2 } }{2}$

==>$\beta =\frac{6-2\sqrt{-1\times2 } }{2}$

==> We know that, i²=-1

==> $\beta =\frac{6-2\sqrt{i^{2} \times2 } }{2}$

==> $\beta =\frac{2(3-\sqrt{i^{2} \times2 }) }{2}$

==> β= 3-i√2

The roots of x²-6x+11 are  3+i√2 and 3-i√2