Question

solving quadratic equations with complex solutions
solve x^2-3x=-8

Answer 1

$x^2-3x=-8\\\\x^2-3x+2.25=-5.75\\\\(x-1.5)^2=-5.75\\x-1.5=\pm i\sqrt{5.75}\\\boxed{x=1.5 \pm i\sqrt{5.75}}$

Not the prettiest but you can simplify

Answer 2

Answer:

$\displaystyle \large{ x = \frac{ 3 \pm \sqrt{ 23}i }{2} }$

Step-by-step explanation:

We are given the equation:-

$\displaystyle \large{ {x}^{2} - 3x = - 8}$

First, arrange the expression to the standard form.

Recall the standard form of Quadratic Equation:-

$\displaystyle \large{a {x}^{2} + bx + c = 0}$

Therefore, add both sides by 8.

$\displaystyle \large{ {x}^{2} - 3x + 8= - 8 + 8} \\ \displaystyle \large{ {x}^{2} - 3x + 8= 0}$

Since we know that the solutions are complex. Therefore, we apply our Quadratic Formula.

Quadratic Formula

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

From the equation, the value of:

• a = 1
• b = -3
• c = 8

Substitute these values in the formula.

$\displaystyle \large{ x = \frac{ - ( - 3)\pm \sqrt{ {( - 3)}^{2} - 4(1)(8)} }{2(1)} } \\ \displaystyle \large{ x = \frac{ 3\pm \sqrt{ 9 -32} }{2} } \\ \displaystyle \large{ x = \frac{ 3 \pm \sqrt{ - 23} }{2} }$

Make sure to recall every necesscary and important fundamental math such as multiplying with negative, exponents, etc.

Imaginary Unit

$\displaystyle \large{i = \sqrt{ - 1} } \\ \displaystyle \large{ {i}^{2} = - 1} \\ \displaystyle \large{ {i}^{3} = - \sqrt{ - 1} = - i} \\ \displaystyle \large{ {i}^{4} = {i}^{2} \times {i}^{2} = ( - 1)( - 1) = 1}$

From √-23, factor √-1 out.

$\displaystyle \large{ x = \frac{ 3 \pm \sqrt{ 23} \sqrt{ - 1} }{2} }$

Convert √-1 to i.

$\displaystyle \large{ x = \frac{ 3 \pm \sqrt{ 23}i }{2} }$

And we're done!