How do you solve #2x²-x+8=0#?

3 Answers
Jun 12, 2018

Answer:

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}#

Jun 12, 2018

Answer:

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#

Jun 12, 2018

Answer:

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.