How do you solve #x^2-2x+9=0#?

Question

Kennedy Adams · Accepted Answer

#x=1+i2sqrt2# or #x=1-i2sqrt2# In the equation #x^2-2x+9=0#, the coefficients are rational but as the discriminant is #(-2)^2-4×1×9=4-36=-32# is negative, the roots of the equation are pair of complex conjugate numbers. Hence, we can use quadratic formula to get roots. Quadratic formula gives the roots of #ax^2+bx+c=0# as #x=(-b+-sqrt(b^2-4ac))/(2a)#. Note that #b^2-4ac# is the discriminant. Hence solution of #x^2-2x+9=0# is given by #x=(-(-2)+-sqrt((-2)^2-4×1×9))/(2×1)# or #x=(2+-sqrt(-32))/2# i.e. #x=(2+i4sqrt2)/2# or #x=(2-i4sqrt2)/2# i.e. #x=1+i2sqrt2# or #x=1-i2sqrt2#

Abigail Allen · Answer

undefined To solve the quadratic equation $x^2 - 2x + 9 = 0$, you can use the quadratic formula:

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

Where $a = 1$, $b = -2$, and $c = 9$.

Plugging these values into the formula:

$$x = \frac{{-(-2) \pm \sqrt{{(-2)^2 - 4 \cdot 1 \cdot 9}}}}{{2 \cdot 1}}$$

$$x = \frac{{2 \pm \sqrt{{4 - 36}}}}{2}$$

$$x = \frac{{2 \pm \sqrt{{-32}}}}{2}$$

Since the discriminant ($b^2 - 4ac$) is negative, the solutions will be complex.

$$x = \frac{{2 \pm \sqrt{{-1 \cdot 32}}}}{2}$$

$$x = \frac{{2 \pm 4i\sqrt{2}}}{2}$$

$$x = 1 \pm 2i\sqrt{2}$$

Therefore, the solutions are $x = 1 + 2i\sqrt{2}$ and $x = 1 - 2i\sqrt{2}$.