Respuesta :

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Answer:

x = (5 + i sqrt(15))/4 or x = (5 - i sqrt(15))/4

Step-by-step explanation:

Solve for x:

2 x^2 - 5 x + 5 = 0

Hint: | Using the quadratic formula, solve for x.

x = (5 ± sqrt((-5)^2 - 4×2×5))/(2×2) = (5 ± sqrt(25 - 40))/4 = (5 ± sqrt(-15))/4:

x = (5 + sqrt(-15))/4 or x = (5 - sqrt(-15))/4

Hint: | Express sqrt(-15) in terms of i.

sqrt(-15) = sqrt(-1) sqrt(15) = i sqrt(15):

Answer: x = (5 + i sqrt(15))/4 or x = (5 - i sqrt(15))/4

The quadratic equation has an imaginary roots, which is can be calculated using the quadratic formula.

What is a quadratic equation ?

Any equation of the form [tex]\rm ax^2+bx+c=0[/tex]  Where x is variable and a, b, and c are any real numbers where a ≠ 0 is called a quadratic equation.

As we know, the formula for the roots of the quadratic equation is given by:

[tex]\rm x = \dfrac{-b \pm\sqrt{b^2-4ac}}{2a}[/tex]

We have a quadratic equation:

2x² - 5x + 5=0

Here a = 2, b = -5, and c = 5

[tex]\rm x = \dfrac{-(-5) \pm\sqrt{(-5)^2-4(2)(5)}}{2(2)}[/tex]

[tex]\rm x = \dfrac{5 \pm\sqrt{-15}}{4}[/tex]

[tex]\rm x = \dfrac{5 \pm i \sqrt{15}}{4}[/tex]  (√(-15) = 15i; i is the iota)

The roots are:

[tex]\rm x = \dfrac{5 + i \sqrt{15}}{4} \\\\\rm x = \dfrac{5 - i \sqrt{15}}{4}[/tex]

Thus, the quadratic equation has an imaginary roots, which is can be calculated using the quadratic formula.

Learn more about quadratic equations here:

brainly.com/question/2263981

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