Courses for Kids
Free study material
Offline Centres
Store Icon

Prove that A.M of the roots of \[{x^2} - 2ax + {b^2} = 0\] is equal to the geometric mean of the roots of the equation \[{x^2} - 2bx + {a^2} = 0\], and vice-versa.

Last updated date: 17th Jul 2024
Total views: 449.4k
Views today: 5.49k
449.4k+ views
Hint: A.M here means arithmetic mean, For this question we have to use the properties of roots of quadratic equations. These properties include the sum of roots and product of roots, equate these properties to the equation given above and then proceed.

Complete step-by-step answer:
We know the standard quadratic equation \[a{x^2} + bx + c = 0\] and let its roots be α and β.
Properties α + β = \[\dfrac{{ - b}}{a}\]……………..(1)
                     α×β = \[\dfrac{c}{a}\]………………….(2)
Taking the equation \[{x^2} - 2ax + {b^2} = 0\], let its roots be m and n
Therefore using equation (1) we get
\[ \Rightarrow m + n = \dfrac{{ - ( - 2a)}}{1}\]
\[ \Rightarrow m + n = 2a\]………..(3)
Since we have to find the arithmetic mean of the equation \[{x^2} - 2ax + {b^2} = 0\]
∴A.M will be \[\dfrac{{m + n}}{2}\] using (3) in it we get
∴\[\dfrac{{2a}}{2}\]= a
∴ A.M of \[{x^2} - 2ax + {b^2} = 0\] is a.
Let the roots of \[{x^2} - 2bx + {a^2} = 0\] be p and q
Now using the property of product of roots i.e. equation (2)
\[ \Rightarrow \]p×q = \[\dfrac{{{a^2}}}{1}\]
\[ \Rightarrow \sqrt {p \times q} = a\]
This shows that A.M of \[{x^2} - 2ax + {b^2} = 0\]= G.M of \[{x^2} - 2bx + {a^2} = 0\]

Note: We know that standard Quadratic equation is \[a{x^2} + bx + c = 0\], where a is the coefficient of\[{x^2}\], b is the coefficient of x and c is the constant and a≠0, since, if a=0, then the equation will no longer remain a quadratic.