Certainly! Let's solve the equation [tex]\( 20x^2 + 10x + 5 = 11x + 6 \)[/tex] step-by-step.
1. Move all terms to one side of the equation:
Start by subtracting [tex]\( 11x \)[/tex] and [tex]\( 6 \)[/tex] from both sides to get the equation in the standard quadratic form [tex]\( ax^2 + bx + c = 0 \)[/tex].
[tex]\[
20x^2 + 10x + 5 - 11x - 6 = 0
\][/tex]
2. Simplify the equation:
Combine like terms:
[tex]\[
20x^2 + (10x - 11x) + (5 - 6) = 0
\][/tex]
[tex]\[
20x^2 - x - 1 = 0
\][/tex]
3. Identify the coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
From the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex], we have:
[tex]\[
a = 20,\quad b = -1,\quad c = -1
\][/tex]
4. Use the quadratic formula:
The quadratic formula is given by:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
5. Calculate the discriminant:
The discriminant ([tex]\( \Delta \)[/tex]) is:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Substitute the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[
\Delta = (-1)^2 - 4 \cdot 20 \cdot (-1)
\][/tex]
[tex]\[
\Delta = 1 + 80
\][/tex]
[tex]\[
\Delta = 81
\][/tex]
6. Find the roots using the quadratic formula:
Substitute [tex]\( \Delta = 81 \)[/tex], [tex]\(a = 20\)[/tex], and [tex]\(b = -1\)[/tex] into the quadratic formula:
[tex]\[
x = \frac{-(-1) \pm \sqrt{81}}{2 \cdot 20}
\][/tex]
[tex]\[
x = \frac{1 \pm 9}{40}
\][/tex]
7. Solve for the two possible values of [tex]\(x\)[/tex]:
- When taking the positive root:
[tex]\[
x = \frac{1 + 9}{40} = \frac{10}{40} = \frac{1}{4}
\][/tex]
- When taking the negative root:
[tex]\[
x = \frac{1 - 9}{40} = \frac{-8}{40} = -\frac{1}{5}
\][/tex]
8. Conclusion:
The solutions to the equation [tex]\( 20x^2 - x - 1 = 0 \)[/tex] are [tex]\( x = \frac{1}{4} \)[/tex] and [tex]\( x = -\frac{1}{5} \)[/tex].
Therefore, the correct option is:
(A) [tex]\( x = -\frac{1}{5}, \frac{1}{4} \)[/tex]
