Question

Find the sum of all possible values of a such that the following equation has real root in x : (x − a)^2 + (x^2−10x+21)^2 = 0?

Answer 1

Answer:

The given equation is :

${(x - a)}^{2} + (x^{2} - 10x + 21) ^{2} = 0$

It is possible only when both the squired terms are individually equal to zero.

That is,

${(x - a)}^{2} = 0 \\ and \\ ({x}^{2} - 10x + 21) ^{2} = 0$

Because it is not possible to have summation of two squared real numbers to be zero.

So from first equation

X=a

And from the second equation

${x}^{2} - 7x - 3x + 21 = 0 \\ x(x - 7) - 3(x - 7) = 0 \\ (x - 3)(x - 7) = 0 \\ x = 3 \\ x = 7 \\ \\ now \: put \: x = 3 \: in \: the \\ equation \: given$

${(3- a)}^{2} + (3^{2} - 10 \times 3 + 21) ^{2} = 0 \\ a = 3 \\ similarly \: put \: x = 7 \\ you \: will \: get \: a= 7$

So the sum of all possible values of a

$a = 3 + 7 = 10$