Given quadratic equation is ,3x^{2}-16x+13=0
the power of the x is 2,so we will get the factors for this equation
when we calculate first term and last term of the equation, 39x^{2}
we rewrite the product in the form of, -13x-3x
3x^{2}-13x-3x+13=0
x(3x-13)-(3x-13)=0
(3x-13)(x-1)=0
=> x=13/3,x=1
SOL--- ,3x2-16x+13=0 . { 13*3 = 39 is the product and the sum is -16 . so the factors are -13 and -3 , because -13*-3 =39 and (-13 ) + (-3) = (-16) }
3x2-16x+13
3x2-13x-3x+13=0
x(3x-13)-1(3x-13)=0
(3x-13)(x-1)=0 => 3x-13 = 0 , x-1 = 0 => 3x=13 , x=1
x=13/3,x=1
3x^{2}-16x+13
= 3x^{2}-3x-13x+13 (Split 16x such that product of those numbers is 39 as a=3 and b=13 also see to it that sum of those number is the middle term i.e -16 in this case.)
= 3x(x-1) -13(x-1)
=(x-1)(3x-13)
therefore,
x-1=0 or 3x-13=0
x=1 or x =13/3
Roots are,
x=1 or x=4.33