To solve the equation [tex]\( 20x^2 + 10x + 5 = 11x + 6 \)[/tex] and determine the correct values for [tex]\( x \)[/tex], we can follow these detailed steps:
1. Move all terms to one side to set the equation to zero:
[tex]\[
20x^2 + 10x + 5 - 11x - 6 = 0
\][/tex]
2. Combine like terms:
[tex]\[
20x^2 + 10x - 11x + 5 - 6 = 0
\][/tex]
[tex]\[
20x^2 - x - 1 = 0
\][/tex]
Now, we have a standard quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = 20 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -1 \)[/tex].
3. Solve the quadratic equation using the quadratic formula:
The quadratic formula is given by:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Substituting the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the formula:
[tex]\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(20)(-1)}}{2(20)}
\][/tex]
Simplify inside the square root:
[tex]\[
x = \frac{1 \pm \sqrt{1 + 80}}{40}
\][/tex]
[tex]\[
x = \frac{1 \pm \sqrt{81}}{40}
\][/tex]
[tex]\[
x = \frac{1 \pm 9}{40}
\][/tex]
4. Calculate the two possible values for [tex]\( x \)[/tex]:
For the positive root:
[tex]\[
x = \frac{1 + 9}{40} = \frac{10}{40} = \frac{1}{4}
\][/tex]
For the negative root:
[tex]\[
x = \frac{1 - 9}{40} = \frac{-8}{40} = -\frac{1}{5}
\][/tex]
So, the solutions to the equation [tex]\( 20x^2 - x - 1 = 0 \)[/tex] are [tex]\( x = -\frac{1}{5} \)[/tex] and [tex]\( x = \frac{1}{4} \)[/tex].
Thus, the correct answer is:
(A) [tex]\( x = -\frac{1}{5}, \frac{1}{4} \)[/tex]
