Chapter ISG, Problem 8.7.2.4P

a.

To determine

To find value of b

Answer to Problem 8.7.2.4P

Value of $b=\frac{-12}{5}$ ,6

Explanation of Solution

Given:

  $25b^2-90b+81=441$

Calculation:

The expression is $25b^2-90b+81=441$

Arranging in standard form,

  $25b^2-90b+81-441=0$

So,

  $25b^2-90b-360=0$ is the standard form.

Dividing the equation by 5,

  $5b^2-18b-72=0$

Factorising the equation,

  $5b^2-30b+12b-72=0$

Taking out common,

  $5b(b-6)+12(b-6)=0$

Taking common again,

  $(5b+12)(b-6)=0$

For expression to be 0,

  $(5b+12)=0$ or $(b-6)=0$

Solving for both,

  $5b=-12$ and $b=6$

  $b=\frac{-12}{5}$ and $b=6$

Hence, value of $b=\frac{-12}{5}$ ,6

b.

To determine

Explain and solve the equation

Answer to Problem 8.7.2.4P

  $x=\frac{-12}{5}$ , $x=6$

Explanation of Solution

Given:

  $25b^2-90b+81=441$

Calculation:

The other way of finding solution of this equation is quadratic formula. The formula is:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ whenand where $a,b,c$ are constants, and $a\ne 0$ .

To calculate solution, find $\sqrt{b^2-4ac}$ ,

  $\sqrt{b^2-4ac}$ = $\sqrt{{(-18)}^2-4(5)(7)}$

Solving,

  = $\sqrt{324-(-1440)}$

  = $\sqrt{1764}$

Substituting the values,

  $x=\frac{18\pm \sqrt{1764}}{2(5)}$

If, $b^2-4ac$ >0, roots are real.

Solving equation,

  $x=6$ ,

  $x=\frac{-12}{5}$