Chapter 9.6, Problem 3BGP

To determine

Find the solution.

Answer to Problem 3BGP

  $x=-1.23$ or $x=2.23$

Explanation of Solution

Given Information: $4x^2-4x-11=0$

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

Calculation:

  $4x^2-4x-11=0$

Now, to make coefficient of $x^2$ = 1, divide the equation by 4.

  $x^2-x-\frac{11}{4}=0$

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

  $x^2-x-\frac{11}{4}+{\left(\frac{1}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2=0$

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

  ${\left(x-\frac{1}{2}\right)}^2-\frac{11}{4}-{\left(\frac{1}{2}\right)}^2=0$

So,

  ${\left(x-\frac{1}{2}\right)}^2-\frac{11}{4}-\frac{1}{4}=0$

Now,

  ${\left(x-\frac{1}{2}\right)}^2-\frac{11-1}{4}=0$

Now,

  ${\left(x-\frac{1}{2}\right)}^2-3=0$

Now, square root of 3 is $\sqrt{3}$

  ${\left(x-\frac{1}{2}\right)}^2-{\left(\sqrt{3}\right)}^2=0$

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

  $\left(x-\frac{1}{2}+\sqrt{3}\right)\left(x-\frac{1}{2}-\sqrt{3}\right)=0$

Now,

  $x=\frac{1}{2}-\sqrt{3}$ or $x=\frac{1}{2}+\sqrt{3}$

Hence,

  $x=-1.23$ or $x=2.23$