Chapter 9.5, Problem 35PPS

To determine

Find the solution.

Answer to Problem 35PPS

$x=0.5$ or $x=0.6$

Explanation of Solution

Given Information: $2x^2-\frac{11}{5}x=-\frac{3}{10}$

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

Calculation:

  $2x^2-\frac{11}{5}x=-\frac{3}{10}$

So, simplify

  $2x^2-\frac{11}{5}x+\frac{3}{10}=0$

Now, to make coefficient of $x^2$ = 1, divide the equation by 2.

  $x^2-\frac{11}{5\times 2}x+\frac{3}{10\times 2}=0$

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

  $x^2-\frac{11}{10}x+\frac{3}{20}+{\left(\frac{11}{20}\right)}^2-{\left(\frac{11}{20}\right)}^2=0$

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

  ${\left(x-\frac{11}{20}\right)}^2+\frac{3}{10}-{\left(\frac{11}{20}\right)}^2=0$

So,

  ${\left(x-\frac{11}{20}\right)}^2+\frac{3}{10}-\frac{121}{400}=0$

Now,

  ${\left(x-\frac{11}{20}\right)}^2+\frac{120-121}{400}=0$

Now,

  ${\left(x-\frac{11}{20}\right)}^2-\frac{1}{400}=0$

Now, square root of 400 is 20 and square root of 1 is 1

  ${\left(x-\frac{11}{20}\right)}^2-{\left(\frac{1}{20}\right)}^2=0$

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

  $\left(x-\frac{11}{20}+\frac{1}{20}\right)\left(x-\frac{11}{20}-\frac{1}{20}\right)=0$

Now,

$x=\frac{11}{20}-\frac{1}{20}$ or $x=\frac{11}{20}+\frac{1}{20}$

Now,

$x=\frac{11-1}{20}$ or $x=\frac{11+1}{20}$

Now,

$x=\frac{10}{20}$ or $x=\frac{12}{20}$

Hence,

$x=0.5$ or $x=0.6$