If one root of the equation ax2 + bx + c be twice the other , show that 2b^2 = 3ac
Answers
Question : -
If one root of the equation ax²+bx+c be twice the other, show that 2b²=9ac ?
ANSWER
Given : -
one root of the equation ax²+bx+c be twice the other
Required to prove : -
- 2b²=3ac
Conditions used : -
Here, conditions refers to relations between the coefficient of the Quadratic equation and their roots.
Relation between the sum of the roots and the coefficients is;
sum of the roots =(- coefficient of x)/(coefficient of x²)
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Relation between the product of the roots and the coefficients is;
product of the roots = (constant term)/(coefficient of x²)
Proof : -
one root of the equation ax²+bx+c be twice the other
let's consider the given quadratic equation as p(x) = ax²+bx+c
The standard form of the quadratic equation is also ax²+bx+c = 0
So,
Here,
- a=a
- b=b
- c=c
Now,
According to given data
one root is twice the other root
So,
Let the one root be a
other root be 2a (since, it is mentioned that one root is twice the other root)
We know that;
Relation between the sum of the roots and the coefficients is;
sum of the roots =(- coefficient of x)/(coefficient of x²)
This implies;
a + 2a = (-b)/(a)
3a =(-b)/(a)
a = (-b)/(a) ÷ (3)/(1)
a = (-b)/(a) x (1)/(3)
a =(-b)/(3a) ...............(1)
consider this as Equation 1
Similarly,
Relation between the product of the roots and the coefficients is;
product of the roots = (constant term)/(coefficient of x²)
a x 2a = (c)/(a)
2a² = (c)/(a)
a² = (c)/(a) ÷ (2)/(1)
a² = (c)/(a) x (1)/(2)
a² = (c)/(2a)
a = √[(c)/(2a)] ......[2]
Since,
The roots are in the condition of one root is twice the other root
Equating both Equation 1 & 2 ( LHS is equal let's equate the RHS part)
(-b)/(3a) = √[(c)/(2a)]
squaring on both sides
{(-b)/(3a)}² = {√[(c)/(2a)]}²
(b²)/(9a²) = (c)/(2a)
b² x 2a = 9a² x c
2ab² = 9a²c
2b² = (9a²c)/(a)
2b² = 9ac
Hence Proved !
Since,
The roots are in the condition of one root is twice the other root
Equating both Equation 1 & 2 ( LHS is equal let's equate the RHS part)
(-b)/(3a) = √[(c)/(2a)]
squaring on both sides
{(-b)/(3a)}² = {√[(c)/(2a)]}²
(b²)/(9a²) = (c)/(2a)
b² x 2a = 9a² x c
2ab² = 9a²c
2b² = (9a²c)/(a)
2b² = 9ac
Hence Proved !