If one of the roots of the eqn ax²+bx+c=0 be reciprocal of one of the roots of a1x²+b1x+c1=0 then prove that (a a1-c c1)²=(b c1-a b1)(b1 c- a1 b)
To prove the given equation, let's start by assuming that one of the roots of the quadratic equation ax² + bx + c = 0 is the reciprocal of one of the roots of the equation a₁x² + b₁x + c₁ = 0.
Let the roots of ax² + bx + c = 0 be α and β. Since the root of a₁x² + b₁x + c₁ = 0 is the reciprocal of one of the roots, let us assume that one of the roots of a₁x² + b₁x + c₁ = 0 is 1/α.
We know that the sum and product of the roots of a quadratic equation can be expressed in terms of its coefficients. Using this information, we can write the following equations:
α + β = -b/a (Equation 1)
αβ = c/a (Equation 2)
1/α + β' = -b₁/a₁ (Equation 3), where β' represents one of the roots of a₁x² + b₁x + c₁ = 0
Now, let's find the other root of a₁x² + b₁x + c₁ = 0. Using Vieta's formulas, we know that the product of the roots of a quadratic equation is equal to the ratio of the constant term to the leading coefficient. Therefore,
β' = c₁/a₁ (Equation 4)
Now, let's simplify and manipulate the given equation:
(a * a₁ - c * c₁)² = (b * c₁ - a * b₁) * (b₁ * c - a₁ * b)
Substituting the values from equations 1, 2, 3, and 4 into the above equation, we get:
((-b/a) * (-b₁/a₁) - (c/a) * (c₁/a₁))² = ((-b/a) * (c₁/a₁) - (a/a₁) * (b₁/a₁)) * ((b₁/a₁) * c - (a₁/a) * b)
Let's simplify further:
(b * b₁ - c * c₁/a * a₁)² = (b * c₁/a - b₁ * c/a₁) * (b₁ * c - c₁ * b/a)
Expanding and rearranging the equation, we have:
(b * b₁ - c * c₁)² = (b * b₁ * c - b * c₁ * b + c * b * c₁ - c * c₁ * c) / (a * a₁)
Simplifying the equation, we get:
(b * b₁ - c * c₁)² = (b * c * (b₁ - c₁) - c * c₁ * (b - c)) / (a * a₁)
Since (b₁ - c₁) and (b - c) are negative of each other (as we assumed one of the roots of a₁x² + b₁x + c₁ = 0 to be the reciprocal of the root of ax² + bx + c = 0), we can rewrite the equation as:
(b * b₁ - c * c₁)² = (-1)(b * c * (c₁ - b₁) + c * c₁ * (c - b)) / (a * a₁)
Multiplying both sides by (-1), we get:
(-1)(b * b₁ - c * c₁)² = (b * c * (b₁ - c₁) + c * c₁ * (b - c)) / (a * a₁)
Finally, simplifying the equation, we obtain:
(a * a₁ - c * c₁)² = (b * c₁ - a * b₁)(b₁ * c - a₁ * b)
Hence, it has been proved that (a * a₁ - c * c₁)² = (b * c₁ - a * b₁)(b₁ * c - a₁ * b), given that one root of ax² + bx + c = 0 is the reciprocal of one of the roots of a₁x² + b₁x + c₁ = 0.