How do you solve #3x² – 16x + 5 = 0#?

2 Answers
Jun 28, 2016

Answer:

#x = 5# or #x = 1/3#

Explanation:

Factorise into #(3x-1)(x -5) = 0#

For the left hand side to be zero, the contents of one of the brackets must evaluate to zero, hence

#3x - 1 = 0 implies x = 1/3#

or

#x - 5 = 0 implies x = 5#.

Alternatively you can use the quadratic formula for equations of the from #ax^2 + bx + c = 0#

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

So in this case, a = 3, b = -16 and c = 5.

#x = (16 + sqrt(256 - 60))/6# or #(16 - sqrt(256 - 60))/6#

#therefore x = 5 or x = 1/3#

Jun 29, 2016

Answer:

#1/3 and 5#

Explanation:

Another way is to use the Transforming Method (Google, Yahoo Search)
#y = 3x^2 - 16x + 5#.
Find the real roots of the transformed equation:
#y' = x^2 - 16x + 15#.
Since a + b + c = 0, use shortcut, the 2 real roots are: X1 = 1 and
#X2 = c/a = 15#.
Back to original equation y, the 2 real roots are: #x1 = (X1)/a = 1/3#
and #x2 = (X2)/a = 15/3 = 5.#