What is the positive solution to the quadratic equation #0=2x^2+3x-8#?

Question

2877 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-08T15:47:43

The positive root #=-3/4+sqrt(73)/4# as an exact value

The positive root #~~1.386# as an approx. value to 3 decimal places

To determine the positive solution find all solutions then filter out the ones you do not want.

Using the standardised formula we have:

It is really is worth while memorising this.

#ax^2+bx+c=0" "# where #a=2"; "b=3"; "c=-8#

Given that:#" "x=(-b+-sqrt(b^2-4ac))/(2a) " "# we have

#=>x=(-3+-sqrt(3^2-4(2)(-8)))/(2(2))#

#x=(-3+-sqrt(9+64))/4#

#x=-3/4+-sqrt(73)/4#

#x=-3/4+-(8.544..)/4#

The positive root #=-3/4+sqrt(73)/4# as an exact value

The positive root #~~1.386# as an approx. value to 3 decimal places