How do you solve #8x^5 + 10x^4 = 4x^3 + 5x^2#?

2 Answers
Mar 10, 2018

Answer:

#x=0 or x=-5/4 or x=+-sqrt(1/2)#

Explanation:

Move all the terms to the left.

#8x^5+10x^4-4x^3-5x^2=0#

Group in pairs and factor each pair.

#(8x^5+10x^4)+(-4x^3-5x^2)=0#

#2x^4(4x+5)-x^2(4x+5)=0#

Factor out the common factor and common bracket:

#x^2(4x+5)(2x^2-1)=0#

There are now three factors. Set each equal to zero and solve.

#x^2 = 0" "rarr x =0#

#4x+5=o" "rarr x =-5/4#

#2x^2-1=0" "rarr x^2 = 1/2" "rarr x = +-sqrt(1/2)#

Mar 10, 2018

Answer:

#x = 0, pm 1/sqrt{2}, -5/4#

Explanation:

We can easily rewrite #8x^5 + 10x^4 = 4x^3 + 5x^2# in the form

# 2x^4(4x+5) = x^2(4x +5)#

Subtracting #x^2(4x +5)# from both sides, leads to

# 2x^4(4x+5) - x^2(4x +5) = 0#

which means that
'
#(2x^4 -x^2)(4x+5)=0#,

i.e.

#x^2(2x^2-1)(4x+5)=0#

Now, for a polynomial to vanish, one of its factors have to be zero. So, either #x^2=0#, or #2x^2-1=0# or #4x+5=0#.

If #x^2=0#, we must have #x=0#

or if #2x^2-1=0#, we get #x^2=1/2#, i.e. #x=pm 1/sqrt{2}#

or, finally, #4x+5=0# which implies that #x=-5/4#