How do you solve the quadratic with complex numbers given $2 x^{2} - 6 x + 7 = 0$ $2x^2-6x+7=0$ ?

Question

2483 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-30T11:17:45

$x = \frac{3}{2} + i \frac{\sqrt{5}}{2}$ $x=3/2+isqrt5/2$ or $x = \frac{3}{2} - i \frac{\sqrt{5}}{2}$ $x=3/2-isqrt5/2$

$2 x^{2} - 6 x + 7 = 0$ $2x^2-6x+7=0$ - we first ty to make complete square as in ${(x + a)}^{2}$ $(x+a)^2$

$2 x^{2} - 6 x + 7 = 0$ $2x^2-6x+7=0$

$\Leftrightarrow x^{2} - 3 x + \frac{7}{2} = 0$ $hArrx^2-3x+7/2=0$

or $x^{2} - 2 \times \frac{3}{2} \times x + {(\frac{3}{2})}^{2} - {(\frac{3}{2})}^{2} + \frac{7}{2} = 0$ $x^2-2xx3/2xx x+(3/2)^2-(3/2)^2+7/2=0$

or ${(x - \frac{3}{2})}^{2} - \frac{9}{4} + \frac{14}{4} = 0$ $(x-3/2)^2-9/4+14/4=0$

or ${(x - \frac{3}{2})}^{2} + \frac{5}{4} = 0$ $(x-3/2)^2+5/4=0$

or ${(x - \frac{3}{2})}^{2} - {(\frac{\sqrt{5}}{2} \times i)}^{2} = 0$ $(x-3/2)^2-(sqrt5/2xxi)^2=0$

As this is of the form $a^{2} - b^{2} = (a + b) (a - b)$ $a^2-b^2=(a+b)(a-b)$

The above is equal to

$(x - \frac{3}{2} - i \frac{\sqrt{5}}{2}) (x - \frac{3}{2} + i \frac{\sqrt{5}}{2}) 0$ $(x-3/2-isqrt5/2)(x-3/2+isqrt5/2)0$

Hence $x - \frac{3}{2} - i \frac{\sqrt{5}}{2} = 0$ $x-3/2-isqrt5/2=0$ or $x - \frac{3}{2} + i \frac{\sqrt{5}}{2} = 0$ $x-3/2+isqrt5/2=0$

i.e. $x = \frac{3}{2} + i \frac{\sqrt{5}}{2}$ $x=3/2+isqrt5/2$ or $x = \frac{3}{2} - i \frac{\sqrt{5}}{2}$ $x=3/2-isqrt5/2$

How do you solve the quadratic with complex numbers given 2x2−6x+7=02x^2-6x+7=0?