Question 940400

find the value of {{{p}}} if the equations 
{{{3x^2-2x+p=0}}} and 
{{{6x^2-17x+12=0}}} 
have a common root 

find roots:

{{{6x^2-17x+12=0}}} 

{{{6x^2-9x-8x+12=0}}}

{{{(6x^2-8x)-(9x-12)=0}}}

{{{2x(3x-4)-3(3x-4)=0}}}

{{{(3x-4)(2x-3)=0}}}.....roots are

{{{(3x-4)=0}}}=>{{{3x=4}}}=>{{{highlight(x=4/3)}}}

and

{{{(2x-3)=0}}}=>{{{2x=3}}}=>{{{x=3/2}}}

{{{3x^2-2x+p=0}}} if {{{x=4/3}}} is same root, then we have

{{{3(4/3)^2-2(4/3)+p=0}}}...solve for {{{p}}}

{{{3(16/9)-2(4/3)+p=0}}}

{{{cross(3)(16/cross(9)3)-(8/3)+p=0}}}

{{{(16/3)-(8/3)+p=0}}}

{{{8/3+p=0}}}

{{{highlight(p=-(8/3))}}}


check the roots using {{{p=-(8/3)}}}

{{{3x^2-2x-(8/3)=0}}}...both sides multiply by {{{3}}}

{{{9x^2-6x-8=0}}}......factor

{{{9x^2-12x+6x-8=0}}}

{{{(9x^2-12x)+(6x-8)=0}}}

{{{3x(3x-4)+2(x-4)=0}}}

{{{(3x+2)(3x-4)=0}}}

roots:

{{{(3x+2)=0}}}=>{{{3x=-2}}}=>{{{x=-2/3}}}

and

{{{(3x-4)=0}}}=>{{{3x=4}}}=>{{{highlight(x=4/3)}}} ...which is same as a root of a given equation {{{6x^2-17x+12=0}}}

let's see it on a graph:


{{{drawing( 600, 600, -15, 15, -15, 15,circle(4/3,0,.2),locate(4/3,0,p(4/3,0)),
 graph( 600, 600, -15, 15, -15, 15, 6x^2-17x+12, 3x^2-2x-(8/3))) }}}