Answer
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Hint:
We must first find out the kind of equations which are given to us. Then we use the concept of that equation to figure out whether that equation will cross the line $y = 0$. If it does, then the number of times or at the number of points at which the equation crosses the line $y = 0$ will be the number of real roots of the equation.
Complete step by step solution:
Let us consider the given equation,
${\left( {x - 1} \right)^2} + {\left( {x - 2} \right)^2} + {\left( {x - 3} \right)^2} + ..... + {\left( {x - n} \right)^2} = 0$
As the terms in the equation are all quadratic in nature, thus the given equation is a quadratic equation.
Now, we also know that the square of any term can never be negative. It can be a 0 or positive value only.
Thus, we can say that the square of the terms given in the equation is always a non-negative value because when we do the square of any number it will always give us a positive value which is always greater than zero.
Thus, the given equation, being quadratic hence having a non-negative value, will never be touch or cross the line $y = 0$, which is the equation of the $x$ -axis. As, this can never happen, thus the equation’s value will always be greater than 0.
As, the left side of the given equation and the right side of the given equation, by the above discussion, can never be equal to each other, thus, we can say that the given equation has no real roots or 0 real roots.
Hence, option (A) is the correct option.
Note:
In this question, you can also expand the terms and combine the like terms. Then you can use the formula for the determinant, ${b^2} - 4ac$, where a, b, and c are the coefficients of ${x^2},x{\text{ and constant.}}$. When you will substitute the values in this formula for determinant, you will observe that the determinant is negative, which implies that there is no real root for the given equation.
We must first find out the kind of equations which are given to us. Then we use the concept of that equation to figure out whether that equation will cross the line $y = 0$. If it does, then the number of times or at the number of points at which the equation crosses the line $y = 0$ will be the number of real roots of the equation.
Complete step by step solution:
Let us consider the given equation,
${\left( {x - 1} \right)^2} + {\left( {x - 2} \right)^2} + {\left( {x - 3} \right)^2} + ..... + {\left( {x - n} \right)^2} = 0$
As the terms in the equation are all quadratic in nature, thus the given equation is a quadratic equation.
Now, we also know that the square of any term can never be negative. It can be a 0 or positive value only.
Thus, we can say that the square of the terms given in the equation is always a non-negative value because when we do the square of any number it will always give us a positive value which is always greater than zero.
Thus, the given equation, being quadratic hence having a non-negative value, will never be touch or cross the line $y = 0$, which is the equation of the $x$ -axis. As, this can never happen, thus the equation’s value will always be greater than 0.
As, the left side of the given equation and the right side of the given equation, by the above discussion, can never be equal to each other, thus, we can say that the given equation has no real roots or 0 real roots.
Hence, option (A) is the correct option.
Note:
In this question, you can also expand the terms and combine the like terms. Then you can use the formula for the determinant, ${b^2} - 4ac$, where a, b, and c are the coefficients of ${x^2},x{\text{ and constant.}}$. When you will substitute the values in this formula for determinant, you will observe that the determinant is negative, which implies that there is no real root for the given equation.
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