Chapter 9.4, Problem 4BGP

To determine

Find the solution.

Answer to Problem 4BGP

  $y=\frac{6}{5}$

Explanation of Solution

Given Information: $25y^2-60y+81=45$

Formula Used: $a^2-2ab+b^2={(a-b)}^2$

Calculation:

  $25y^2-60y+81=45$

So,

  $25y^2-60y+81-45=0$

Now, simplify

  $25y^2-60y+36=0$

Divide the equation by coefficient of $y^2$ i.e. 25.

  $y^2-\frac{60}{25}y+\frac{36}{25}=0$

Now, simplify

  $y^2-\frac{12}{5}y+\frac{36}{25}=0$

Now, adding & subtracting the square of half the coefficient of y.

  $y^2-\frac{12}{5}y+\frac{36}{25}+{\left(\frac{6}{5}\right)}^2-{\left(\frac{6}{5}\right)}^2=0$

So,

  $y^2-\frac{12}{5}y+\frac{36}{25}+\frac{36}{25}-\frac{36}{25}=0$

Now, simplify

  $y^2-\frac{12}{5}y+\frac{36}{25}=0$

Now, square root of 36 is 6 and square root of 25 is 5.

  $y^2-\frac{12}{5}y+{\left(\frac{6}{5}\right)}^2=0$

Now use formula, $a^2-2ab+b^2={(a-b)}^2$

  ${\left(y-\frac{6}{5}\right)}^2=0$

Now, square root both sides.

  $y-\frac{6}{5}=0$

Hence,

  $y=\frac{6}{5}$