Solve {left(sqrt{50}right)}^2+left(x-1right)^2=x^2

$${ \left( \sqrt{ 50 } \right) }^{ 2 } + { \left(x-1 \right) }^{ 2 } = { x }^{ 2 }$$

Solve for x

$$x = \frac{51}{2} = 25\frac{1}{2} = 25.5$$

Steps for Solving Linear Equation

The square of \sqrt{50} is 50.

$$50+\left(x-1\right)^{2}=x^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$50+x^{2}-2x+1=x^{2}$$

Add 50 and 1 to get 51.

$$51+x^{2}-2x=x^{2}$$

Subtract x^{2} from both sides.

$$51+x^{2}-2x-x^{2}=0$$

Combine x^{2} and -x^{2} to get 0.

$$51-2x=0$$

Subtract 51 from both sides. Anything subtracted from zero gives its negation.

$$-2x=-51$$

Divide both sides by -2.

$$x=\frac{-51}{-2}$$

Fraction \frac{-51}{-2} can be simplified to \frac{51}{2} by removing the negative sign from both the numerator and the denominator.

$$x=\frac{51}{2}$$