Find the roots of the following quadratic equation if they exist by the method of completing the square 2x²-7x+3=0?

2 Answers
Jun 23, 2018

See my proof below.

Explanation:

Completing the square:

#x^2-2*7/4x+49/16+3/2-49/16=0#
this is

#(x-7/4)^2-25/16=0#
using that

#(a^2-b^2=(a-b)(a+b)#
then we get

#(x-7/4+5/4)(x-7/4-5/4)=0#

#(x-1/2)(x-3)=0#

Jun 23, 2018

#x=1/2# and #x=3#

Explanation:

It is always good to write out the square as a reminder. We want the expression on the form:
#(x-a)^2=b#
#2(x-a)^2=2(x^2-2ax+a^2)=2x^2-7x=-3#
gives
#(x-a)^2=(x^2-2ax+a^2)=x^2-7/2x+(7/4)^2#
=#-3/2+(7/4)^2=(-24+49)/16=25/16#
where I have moved the constant over to the right side.

Comparing term for term we see that
#2a=7/2# or #a=7/4#
The square, therefore, is
#(x-7/4)^2=25/16#

We, therefore get:
=#x-7/4=+-5/4#

Therefore:
#x=7/4+-5/4#
This gives the two solutions #x=1/2# and #x=3#

Test:
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We see the graph of #f(x)=2x^(2)-7x+3# crosses the x axis in x=0.5 and x=3