How do you solve #x= - 4x ^ { 2} + 5#?

1 Answer
Jim G.
Oct 3, 2017

#x=-5/4" or "x=1#

Explanation:

#"rearrange terms into standard form "color(white)(x)ax^2+bx+c=0#

#rArr4x^2+x-5=0#

#"the factors of - 20 which sum to + 1 are + 5 and - 4"#

#rArr4x^2-4x+5x-5=0larrcolor(blue)" split middle term"#

#"factorise by grouping"#

#color(red)(4x)(x-1)color(red)(+5)(x-1)=0#

#"take out "color(blue)"common factor of "(x-1)#

#rArr(x-1)(color(red)(4x+5))=0#

#"equate each factor to zero and solve for x"#

#4x+5=0rArrx=-5/4#

#x-1=0rArrx=1#

Related questions
Impact of this question
515 views around the world
You can reuse this answer
Creative Commons License