How do you solve #9p ^ { 2} - 24p + 11= 0#?

2 Answers
marfre
Mar 31, 2017

Use the quadratic formula to get #x = 4/3 +- sqrt(5)/3#

Explanation:

Use the quadratic formula #Ax^2 + Bx + C = 0#

#x =( -B+-sqrt(B^2-4AC))/(2A) = (24 +- sqrt(576 - 4(9)(11)))/(18)#

#x = 4/3 +- sqrt(180)/18 = 4/3 +- sqrt(36 * 5)/18 = 4/3 +- (6sqrt(5))/18#

#x = 4/3 +- sqrt(5)/3#

George C.
Mar 31, 2017

#p = 4/3+-sqrt(5)/3#

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

We will use this with #a=(3p-4)# and #b=sqrt(5)#.

Given:

#9p^2-24p+11 = 0#

Notice that #9p^2 = (3p)^2# and #24p = 2(3p)(4)#

So consider:

#(3p-4)^2 = (3p)^2-2(3p)(4)+4^2#

#color(white)((3p-4)^2) = 9p^2-24p+16#

So:

#0 = 9p^2-24p+11#

#color(white)(0) = 9p^2-24p+16-5#

#color(white)(0) = (3p-4)^2-(sqrt(5))^2#

#color(white)(0) = ((3p-4)-sqrt(5))((3p-4)+sqrt(5))#

#color(white)(0) = (3p-4-sqrt(5))(3p-4+sqrt(5))#

Hence:

#3p = 4+-sqrt(5)#

So:

#p = 4/3+-sqrt(5)/3#

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