If AM and GM of roots of a quadratic equation are 6 and 4 respectively, then find out the quadratic equation and solve it.

Let the root of quadratic eqn are a and b As given in question AM=(a+b)/2=6 =>a+b=12-----------------(eqn 1) GM=square root of (ab)=4 =>ab=16--------------(enq 2) As you know quadratic eqn is given by:- X^2-(sum of roots)X+product of roots=0 i.e X^2-(a+b)X+ab=0 from eqn 1 and egn 2 X^2-12x+16=0 this is the required eqn... read less

If A and B are roots of a quadratic equation, then the equation can be written as follow:- x^2 - (A+B)x + (A*B)=0. It is given that the arithmetic mean of the roots is 6. This means... (A+B)/2 =6 further implying (A+B) =12. Also the geometric mean of the roots is given as 4. this means... square root of (A*B)=4, Therefore A*B =16. Hence, the quadratic equation is x^2-12x+16=0. Roots of a quadratic equation of the form a(x^2)+bx+c=0 is given by x= [-b+ sqrt(b^2)-4ac)]/2a and [-b- sqrt(b^2)-4ac)]/2a. Here a=1, b= -12 and c=16. putting this we get, x= [12 + sqrt(144-64)]/2 and [12 - sqrt(144-64)]/2, = [12 + sqrt(80)]/2 and [12 - sqrt(80)]/2 = [12 + sqrt(16*5)]/2 and [12 + sqrt(16*5)]/2 taking 16 outside the sqrt sign = [12 + 4sqrt(5)]/2 and [12 - 4sqrt(5)]/2 now simplify by dividing each term of numerator (separated by + sign) by denominator x=6+2sqrt(5) and 6-2sqrt(5). NOTE:- sqrt here refers to operation of taking square root. read less