How do I use the quadratic formula to solve #3x^2 - 6 = 4x#?

1 Answer
Mar 21, 2018

Answer:

#x = 2/3+-sqrt(22)/3#

Explanation:

Given:

#3x^2-6 = 4x#

Subtract #4x# from both sides to get:

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

This is now in standard form:

#ax^2+bx+c = 0#

with #a=3#, #b=-4# and #c=-6#

So we can apply the quadratic formula to find:

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

#color(white)(x) = (-(color(blue)(-4))+-sqrt((color(blue)(-4))^2-4(color(blue)(3))(color(blue)(-6))))/(2(color(blue)(3)))#

#color(white)(x) = (4+-sqrt(16+72))/6#

#color(white)(x) = (4+-sqrt(88))/6#

#color(white)(x) = (4+-sqrt(2^2 * 22))/6#

#color(white)(x) = (4+-2sqrt(22))/6#

#color(white)(x) = (2+-sqrt(22))/3#

#color(white)(x) = 2/3+-sqrt(22)/3#

Bonus

Notice in the derivation above that we had a factor #2^2# in the radicand that we brought outside the square root as #2# and then cancelled with the #2# in the divisor.

This always happens when #b# is even, adding an extra simplification step or two.

In such cases we can instead think of the original quadratic as taking the form:

#ax^2+2dx+c = 0#

Then the solutions can be expressed as:

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

In our example #a=3#, #d=-2# and #c=-6#, so we get:

#x = (-(color(blue)(-2))+-sqrt((color(blue)(-2))^2-(color(blue)(3))(color(blue)(-6))))/(color(blue)(3))#

#color(white)(x) = (2+-sqrt(4+18))/3#

#color(white)(x) = (2+-sqrt(22))/3#

#color(white)(x) = 2/3+-sqrt(22)/3#

I don't know how widely this variant of the quadratic formula is used.

If #a=1# then we can simplify further to say that the solutions of:

#x^2+2dx+c = 0#

are:

#x = -d+-sqrt(d^2-4c)#