How do you solve #3p ^ { 2} + 2= 6p#?

1 Answer
BDT Stu.
Dec 4, 2017

#1/3(3+-sqrt(3))# . Get this answer by completing the square and making #p# the subject of the equation.

Explanation:

Rearrange the formula by bringing all values to one side of the equation.

#3p^2-6p+2=0#

Leave the #2# in the formula but still ignore it and do simple factorisation on the rest.

#3[p^2-2p]+2=0#

Now complete the square in the brackets like you would do when you are working with quadratic formulas with the coefficient of #1#. In this equation the coefficient is #3# so it has to be factorised away.

#3[(p-1)^2-1]+2=0#

Expand the square brackets only because we want the equation to remain completed.

#3(p-1)^2-3+2=0#

Simplify by adding like term.

#3(p-1)^2-1=0#

Rearrange the formula to make #p# the subject.

#3(p-1)^2=1#
#(p-1)^2=1/3#
#p-1=+-sqrt(1/3)#
#p=1+-sqrt(1/3)#

Tip! In the equation #ax^2+bx+c#
the formula for completing the square is

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Rationalize the equation.

#p=1+-sqrt(1)/sqrt(3)#
#p=(1*sqrt(3))/sqrt(3)+-sqrt(1)/sqrt(3)#
#p=sqrt(3)/sqrt(3)+-sqrt(1)/sqrt(3)#
#p=(sqrt(3)+-sqrt(1))/sqrt(3)#
#p=(sqrt(3)+-sqrt(1))/sqrt(3)*sqrt(3)/sqrt(3)#
#p=(sqrt(3)(sqrt(3)+-sqrt(1)))/(sqrt(3)*sqrt(3))#
#p=(sqrt(3)*sqrt(3)+-(sqrt(3)*sqrt(1)))/(sqrt(3)*sqrt(3))#
#p=(3+-sqrt(3))/3#
#p=1/3(3+-sqrt(3))#

Learn about rationalising at the following link
https://www.mathsisfun.com/algebra/rationalize-denominator.html

IF an exam question asks for two answers, write one answer with #+# sign instead of #+-# and the other with #-# sign.

In calculator exam questions, write the two answers in number form, but not surd.

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