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Find the solutions to the equation below. Check all that apply.

[tex]\[ 20x^2 - 26x + 8 = 0 \][/tex]

A. [tex]\( x = \frac{4}{5} \)[/tex]

B. [tex]\( x = \frac{1}{2} \)[/tex]

C. [tex]\( x = \frac{3}{5} \)[/tex]

D. [tex]\( x = \frac{1}{3} \)[/tex]

E. [tex]\( x = \frac{2}{3} \)[/tex]

F. [tex]\( x = \frac{1}{5} \)[/tex]


Sagot :

To find the solutions to the quadratic equation:
[tex]\[ 20x^2 - 26x + 8 = 0, \][/tex]
we need to solve it and then verify which, if any, of the given choices are correct solutions.

### Step 1: Solving the Quadratic Equation
The given quadratic equation is:
[tex]\[ 20x^2 - 26x + 8 = 0. \][/tex]

Using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \][/tex]
where [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are coefficients from the equation [tex]\(ax^2 + bx + c = 0\)[/tex].

Here, [tex]\(a = 20\)[/tex], [tex]\(b = -26\)[/tex], and [tex]\(c = 8\)[/tex].

First, calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac. \][/tex]
[tex]\[ \Delta = (-26)^2 - 4(20)(8). \][/tex]
[tex]\[ \Delta = 676 - 640. \][/tex]
[tex]\[ \Delta = 36. \][/tex]

Since the discriminant is positive, there are two real solutions:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a}. \][/tex]
[tex]\[ x = \frac{26 \pm \sqrt{36}}{40}. \][/tex]
[tex]\[ x = \frac{26 \pm 6}{40}. \][/tex]

This gives us two solutions:
[tex]\[ x_1 = \frac{26 + 6}{40} = \frac{32}{40} = \frac{4}{5}, \][/tex]
[tex]\[ x_2 = \frac{26 - 6}{40} = \frac{20}{40} = \frac{1}{2}. \][/tex]

The solutions are:
[tex]\[ x = \frac{4}{5} \quad \text{and} \quad x = \frac{1}{2}. \][/tex]

### Step 2: Verifying Each Choice
Let's check each of the given choices to see if they satisfy the equation [tex]\(20x^2 - 26x + 8 = 0\)[/tex].

A. [tex]\( x = \frac{4}{5} \)[/tex]
[tex]\[ 20 \left( \frac{4}{5} \right)^2 - 26 \left( \frac{4}{5} \right) + 8 = 0, \][/tex]
[tex]\[ 20 \left( \frac{16}{25} \right) - 26 \left( \frac{4}{5} \right) + 8 = 0, \][/tex]
[tex]\[ \frac{320}{25} - \frac{104}{5} + 8 = 0, \][/tex]
[tex]\[ 12.8 - 20.8 + 8 = 0, \][/tex]
[tex]\[ 0 = 0. \][/tex]
This is a correct solution.

B. [tex]\( x = \frac{1}{2} \)[/tex]
[tex]\[ 20 \left( \frac{1}{2} \right)^2 - 26 \left( \frac{1}{2} \right) + 8 = 0, \][/tex]
[tex]\[ 20 \left( \frac{1}{4} \right) - 13 + 8 = 0, \][/tex]
[tex]\[ 5 - 13 + 8 = 0, \][/tex]
[tex]\[ 0 = 0. \][/tex]
This is a correct solution.

C. [tex]\( x = \frac{3}{5} \)[/tex]
[tex]\[ 20 \left( \frac{3}{5} \right)^2 - 26 \left( \frac{3}{5} \right) + 8 \][/tex]
[tex]\[ 20 \left( \frac{9}{25} \right) - 26 \left( \frac{3}{5} \right) + 8 \][/tex]
[tex]\[ \frac{180}{25} - \frac{78}{5} + 8 \][/tex]
[tex]\[ 7.2 - 15.6 + 8 \][/tex]
[tex]\[ -0.4 \neq 0 \][/tex]
This is not a solution.

D. [tex]\( x = \frac{1}{3} \)[/tex]
[tex]\[ 20 \left( \frac{1}{3} \right)^2 - 26 \left( \frac{1}{3} \right) + 8 \][/tex]
[tex]\[ 20 \left( \frac{1}{9} \right) - 26 \left( \frac{1}{3} \right) + 8 \][/tex]
[tex]\[ \frac{20}{9} - \frac{26}{3} + 8 \][/tex]
[tex]\[ \frac{20 - 78 + 72}{9} \][/tex]
[tex]\[ \frac{14}{9} \neq 0 \][/tex]
This is not a solution.

E. [tex]\( x = \frac{2}{3} \)[/tex]
[tex]\[ 20 \left( \frac{2}{3} \right)^2 - 26 \left( \frac{2}{3} \right) + 8 \][/tex]
[tex]\[ 20 \left( \frac{4}{9} \right) - 26 \left( \frac{2}{3} \right) + 8 \][/tex]
[tex]\[ \frac{80}{9} - \frac{52}{3} + 8 \][/tex]
[tex]\[ \frac{80 - 156 + 72}{9} \][/tex]
[tex]\[ \frac{-4}{9} \neq 0 \][/tex]
This is not a solution.

F. [tex]\( x = \frac{1}{5} \)[/tex]
[tex]\[ 20 \left( \frac{1}{5} \right)^2 - 26 \left( \frac{1}{5} \right) + 8 \][/tex]
[tex]\[ 20 \left( \frac{1}{25} \right) - 26 \left( \frac{1}{5} \right) + 8 \][/tex]
[tex]\[ \frac{20}{25} - \frac{26}{5} + 8 \][/tex]
[tex]\[ 0.8 - 5.2 + 8 \][/tex]
[tex]\[ 0 \neq 0 \][/tex]
This is not a solution.

### Conclusion
The correct solutions to the equation [tex]\(20x^2 - 26x + 8 = 0\)[/tex] from the given choices are:
- [tex]\( x = \frac{4}{5} \)[/tex] [tex]\((A)\)[/tex]
- [tex]\( x = \frac{1}{2} \)[/tex] [tex]\((B)\)[/tex]

So, the correct choices are:
- A. [tex]\( x = \frac{4}{5} \)[/tex]
- B. [tex]\( x = \frac{1}{2} \)[/tex]