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

2 Answers
Mar 22, 2018

Answer:

The solutions to the equation are #x=1/2# and #x=3/2#.

Explanation:

We know that this is a quadratic since it has an #x^2# term in it.

To solve quadratics, first set one side equal to #0#:

#4x^2+3=8x#

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

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

Now, we have to find two numbers that multiply to #12# (the #a# term times the #c# term) and add up to #-8# (the #b# term).

After some experimentation, you will find that these two numbers are #-6# and #-2#.

Now, split up the #x# terms into these two numbers, then factor the first two terms and the last two terms separately, and lastly, combine them:

#4x^2-6x-2x+3=0#

#color(red)(2x)(2x-3)-2x+3=0#

#color(red)(2x)(2x-3)-color(blue)1(2x-3)=0#

#(color(red)(2x)-color(blue)1)(2x-3)=0#

Now, solve for when each factor equals #0#, and those will be the solutions:

#color(white){color(black)( (2x-1=0,qquadqquad2x-3=0), (2x=1,qquadqquad2x=3), (x=1/2,qquadqquadx=3/2):}#

Those are the answers. We can check out answers by graphing the quadratic #4x^2-8x+3# on a calculator and seeing where it crosses the #x#-axis (it should cross when #x=1/2# and also when #x=3/2#):

graph{4x^2-8x+3 [-0.5, 2.5, -1.29, 0.419]}

It does, so our answers are correct. Hope this helped!

Mar 22, 2018

Answer:

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

Explanation:

Solve the equation by the new Transforming Method (Socratic Search).

#y = 4x^2 - 8x + 3 = 0#

Transformed equation:

#y' = x^2 - 8x + 12#.

Proceeding. Find #2# real roots of #y'#, then, divide them by #a = 4#.

Find #2# real roots knowing the sum #(-b = 8)# and the product
#(ac = 12)#. They are: #2# and #6#.

The #2# real roots of #y# are:

#x_1 = 2/a = 2/4 = color(blue)(1/2)# and #x_2 = 6/a = 6/4 = color(blue)(3/2)#.