Solve this quadratic equation. Return the answer in 2 decimals ?

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3 Answers
Feb 16, 2018

Answer:

#x=3.64,-0.14#

Explanation:

We have #2x-1/x=7#

Multiplying both sides by #x#, we get:

#x(2x-1/x)=7x#

#2x^2-1=7x#

#2x^2-7x-1=0#

Now we have a quadratic equation. For any #ax^2+bx+c=0#, where #a!=0,# #x=(-b+-sqrt(b^2-4ac))/(2a)#.

Here, #a=2,b=-7,c=-1#

We can input:

#(-(-7)+-sqrt((-7)^2-4*2*-1))/(2*2)#

#(7+-sqrt(49+8))/4#

#(7+-sqrt(57))/4#

#x=(7+sqrt(57))/4,(7-sqrt(57))/4#

#x=3.64,-0.14#

Feb 16, 2018

Answer:

#x = 3.64 or x = -0.14#

Explanation:

This is clearly not a comfortable form to work with.
Multiply through by #x# and re-arrange the equation into the form:

#ax^2 +bx+c=0#

#2xcolor(blue)(xx x) -1/xcolor(blue)(xx x) =7color(blue)(xx x)#

#2x^2 -1=7x#

#2x^2 -7x-1=0" "larr# it does not factorise

#x= (-b+-sqrt(b^2-4ac))/(2a)#

#x = (-(-7)+-sqrt((-7)^2 -4(2)(-1)))/(2(2))#

#x = (7+-sqrt(49+8))/(4)#

#x = (7+sqrt57)/4 = 3.64#

#x = (7-sqrt57)/4 = -0.14#

Feb 16, 2018

Answer:

See below...

Explanation:

First we need the standard format of #ax^2+bx+c=0#

First we multiply all by #x# to remove the fraction.

#2x-1/x=7 => 2x^2-1=7x#

Now we move the #7x# over by subtracting both sides by #7x#

#2x^2-1=7x => 2x^2-7x-1=0#

As we want the answers to #2d.p# it strongly hints that we need to use the quadratic formula.

We know that #x=-b+-sqrt(b^2-4ac)/(2a)#

Now from our equation we know that ...

#a =2#, #b=-7# and #c=-1#

Now we plug these into our formula, but as we have a #+# and a #-# we have to do it twice.

#x=-(-7)+sqrt((-7)^2-4(2)(-1))/(2(2))#
#x=-(-7)-sqrt((-7)^2-4(2)(-1))/(2(2))#

Now we put each one into our calculator and round to #2d.p.#

#therefore x =-0.14 , x =3.64 #

Both to #2d.p#