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Class 12 Tuition > Learn Class XI-XII Tuition (PUC) > If AM and GM of roots of a quadratic equation are 6...
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If AM and GM of roots of a quadratic equation are 6 and 4 respectively, then find out the quadratic equation and solve it.

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x2-12x+16
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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...
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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
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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...
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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
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SME in Mathematics & Statistics

A quadratic equation whose roots are r1 and r2, is x2 - (r1 + r2)x + r1r2 = 0 Given, (r1 + r2)/2 = 6 and sqrt(r1r2) = 4 Hence, the quadratic equation is x2 - 12x + 16 = 0 Roots are 6+2?5, 6-2?5
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Math Educator for Std.11th ,12th , Engineering Entrance and Degree Level with 11+ Years Experience

Let 'a' and 'b' be the roots of the quadratic equation. A.M. of a, b = a + b / 2 = 6 , G.M. of a , b = ab = square4 = 16 ,therefore, a + b = 12 , ab = 16 . Equation whose roots are a , b is x2 - (a + b)x + ab = 0 , then required quadratic equation is x2 - 12x + 16 = 0
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Put The AM as b and GM as c in standard form.
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Online Tutor-Mathematics, Science, Competitive Exam Preparation

6+2?5, 6-2?5 are the roots Quadratic equation is x2-12x-16=0 Hint Sum of roots = -b/a = AM*2 Product of roots = c/a = GM^2
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The quadratic equation is x² - 12x +16 = 0, and the roots are (6 - 2?5) and (6 + 2?5)
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Let, roots are a and b. So Equation is : (x-a)(x-b)=0 => x^2 --a.x-b.x +a.b=0=> x^2 --(a+b)x + a.b=0---- (1) According to the question, AM=(a+b)/2= 6=> a+b=12 GM=?(a.b)=4 => a.b= 16. Substituting these in (1), The required quadratic equation is: x^2 --12x+16=0, To solve the equation, use Shridharacharya’s...
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Let, roots are a and b. So Equation is : (x-a)(x-b)=0 => x^2 –a.x-b.x +a.b=0=> x^2 –(a+b)x + a.b=0---- (1) According to the question, AM=(a+b)/2= 6=> a+b=12 GM=?(a.b)=4 => a.b= 16. Substituting these in (1), The required quadratic equation is: x^2 –12x+16=0, To solve the equation, use Shridharacharya’s Formula read less
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x^2 -12x +16 =0
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