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.
Answer
361.5k+ 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.
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.
Last updated date: 27th Sep 2023
•
Total views: 361.5k
•
Views today: 9.61k
Recently Updated Pages
What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Why are resources distributed unequally over the e class 7 social science CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Briefly mention the contribution of TH Morgan in g class 12 biology CBSE

What is the past tense of read class 10 english CBSE
