
If $\alpha$,$\beta$ are the roots of $(x-a)(x-b)=c$, $c\neq0$, then the roots of $(x-\alpha)(x-\beta)+c=0$ shall be
A) $a, c$
B) $b,c$
C) $a,b$
D) $a+c,b+c$
Answer
163.8k+ views
Hint: To solve this problem, we will use the properties of roots of a quadratic equation. By understanding the relationship between the roots and coefficients of a quadratic equation, we can simplify the given expressions and find the correct answer.
Formula Used:(x-a)(x-b)=c, where a and b are roots of the equation and c is a non-zero constant.
Complete step by step solution:We can start by understanding that a quadratic equation can be factored into the product of two linear factors, (x−a)(x−b) and the constant c represents the coefficient of the x^2 term.
Given the equation (x−a)(x−b)=c, we can see that the roots of this equation are a and b.
Now, if we add c to both sides of the equation, we get (x−a)(x−b)+c=c+c, which can be rewritten as (x−a)(x−b)+c=0.
This new equation can be factored as (x−a)(x−b)+c=0 and the roots of this equation are a+c and b+c.
We know that the roots of a quadratic equation of the form $(x-a)(x-b)=c$ are given by a and b.
Given that $(x-\alpha)(x-\beta)+c=0$ where a and b are the roots of the equation and c is a non-zero constant.
$x^2-(\alpha+\beta)x+(\alpha\beta+c)=0$
Comparing the coefficients we get
$\alpha+\beta=a+b$
$\alpha\beta+c=ab$
$\alpha,\beta$ are the roots of the equation $x^2-(a+b)x+(ab+c)$
As we are given $c\neq 0$ which means the roots are distinct, so these roots are $a+c,b+c$
Option ‘D’ is correct
Note: The key to solving this problem is understanding the relationship between the roots and coefficients of a quadratic equation. By comparing the coefficients we get the roots.
Formula Used:(x-a)(x-b)=c, where a and b are roots of the equation and c is a non-zero constant.
Complete step by step solution:We can start by understanding that a quadratic equation can be factored into the product of two linear factors, (x−a)(x−b) and the constant c represents the coefficient of the x^2 term.
Given the equation (x−a)(x−b)=c, we can see that the roots of this equation are a and b.
Now, if we add c to both sides of the equation, we get (x−a)(x−b)+c=c+c, which can be rewritten as (x−a)(x−b)+c=0.
This new equation can be factored as (x−a)(x−b)+c=0 and the roots of this equation are a+c and b+c.
We know that the roots of a quadratic equation of the form $(x-a)(x-b)=c$ are given by a and b.
Given that $(x-\alpha)(x-\beta)+c=0$ where a and b are the roots of the equation and c is a non-zero constant.
$x^2-(\alpha+\beta)x+(\alpha\beta+c)=0$
Comparing the coefficients we get
$\alpha+\beta=a+b$
$\alpha\beta+c=ab$
$\alpha,\beta$ are the roots of the equation $x^2-(a+b)x+(ab+c)$
As we are given $c\neq 0$ which means the roots are distinct, so these roots are $a+c,b+c$
Option ‘D’ is correct
Note: The key to solving this problem is understanding the relationship between the roots and coefficients of a quadratic equation. By comparing the coefficients we get the roots.
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