Question

Find the value of r. (r-5)^2=0

Answer 1

$( {r - 5)}^{2} = {r}^{2} + ( {5)}^{2} - 2(r)(5) \\ = {r}^{2} + 25 - 10r$

${r}^{2} - 10r + 25 \\ {r}^{2} - 5r - 5r + 25 \\ {r}(r - 5) - 5(r - 5) = 0 \\ (r - 5)(r - 5) = 0 \\ r - 5 = 0 \\ r = 5$

Answer 2

Answer:-

• The value of r = 5.

Solution:-

$\sf \implies \: {(r - 5)}^{2} = 0 \\ \\ \sf \implies \: {r}^{2} - 25 = 0 \\ \\ \sf \implies \: {r}^{2} = 25 \\ \\ \sf \implies \: r = \sqrt{25} \\ \\ \sf \implies \: r = 5$