Question

How do you solve `2x²-x+8=0`?

Answer 1

See a solution process below:

Explanation:

We can use the quadratic equation to solve this problem:

The quadratic formula states:

For `color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0`, the values of `x` which are the solutions to the equation are given by:

`x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))`

Substituting:

`color(red)(2)` for `color(red)(a)`

`color(blue)(-1)` for `color(blue)(b)`

`color(green)(8)` for `color(green)(c)` gives:

`x = (-color(blue)(-1) +- sqrt(color(blue)(-1)^2 - (4 * color(red)(2) * color(green)(8))))/(2 * color(red)(2))`

`x = (1 +- sqrt(1 - 64))/4`

`x = (1 +- sqrt(-63))/4`

`x = (1 +- sqrt(63 * -1))/4`

`x = (1 +- sqrt(63)sqrt(-1))/4`

The square root of -1 or `sqrt(-1)` is known as an imaginary number and is often referenced as `i`.

Substituting gives the solution set as;

`x = {(1 - sqrt(63)i)/4, (1 +- sqrt(63)i)/4}`

Answer 2

No real roots

Explanation:

`y = 2x^2 - x + 8 = 0`
Use the improved quadratic formula (Socratic, Google Search).
`D = d^2 = b^2 - 4ac = 1 - 64 = -63`
Since D < 0, there are no real roots.
There are 2 complex roots:
`x = -b/(2a) +- d/(2a)`, with `d = +- isqrt63`
`x = 1/4 +- (isqrt63)/4 = (1 +- isqrt63)/4`

Answer 3

See below..

Explanation:

Let's complete the square:

`2[x^2-1/2x]+8:`

`rArr 2(x-1/4)^2-1/16+128/16`

`rArr2(x-1/4)^2+127/16`

`rArr 2(x-1/4)^2=-127/16`

`rArr (x-1/4)^2=-127/32`

`x-1/4=pmsqrt(-127/32)`

`x=1/4pmsqrt(-127/32)`

No real roots as negative square root, although you could use `i`, but I don't know how.