Question

for real no x and y ,x^2+(y-4)^2=0 then find the value of (x+y)​

Answer 1

$\begin{gathered}\begin{gathered}\bf \:Given-\begin{cases} &\sf{Two \: real \: numbers \: x \: and \: y \: such \: that} \\ &\sf{ {x}^{2} + {(y - 4)}^{2} = 0} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To \: Find - \begin{cases} &\sf{value \: of \: x + y}\end{cases}\end{gathered}\end{gathered}$

$\large\underline{\bf{Solution-}}$

↝ Given

$\rm :\longmapsto\: {x}^{2} + {(y - 4)}^{2} = 0$

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ The sum of squares is 0, its possible only when number is itself zero.

$\rm :\longmapsto\:x = 0 \: \: \: and \: \: \: y - 4 = 0$

$\bf\implies \:x = 0 \: \: \: and \: \: \: y = 4$

So,

$\rm :\longmapsto\:x + y = 0 + 4 = 4$

$\overbrace{ \underline { \boxed { \bf \therefore \: The \:value \: of \: (x + y)\: is \: 4}}}$

Additional Information :-

Real numbers can be positive or negative, and include the number zero. They are called real numbers because they are not imaginary, which is a different system of numbers. Imaginary numbers are numbers that cannot be quantified, like the square root of -1.

• The real numbers include integers, rational, and irrational numbers.

• The number line contains all the real numbers and nothing else.

• Every real number has a decimal representation.Real numbers can do arithmetic.

• There are Numbers that are Not Real (imaginary, complex).