Certainly! Let's solve the quadratic equation [tex]\(x^2 - 0.7x + 0.1 = 0\)[/tex] step by step.
A general quadratic equation is of the form:
[tex]\[ ax^2 + bx + c = 0 \][/tex]
where [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are constants.
In our specific equation, we identify:
[tex]\[ a = 1, \; b = -0.7, \; c = 0.1 \][/tex]
We can solve this using the quadratic formula, which is given by:
[tex]\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \][/tex]
Now we will substitute the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] into the quadratic formula:
1. Compute the discriminant:
[tex]\[ \text{Discriminant} = b^2 - 4ac \][/tex]
[tex]\[ \text{Discriminant} = (-0.7)^2 - 4 \cdot 1 \cdot 0.1 \][/tex]
[tex]\[ \text{Discriminant} = 0.49 - 0.4 \][/tex]
[tex]\[ \text{Discriminant} = 0.09 \][/tex]
2. Take the square root of the discriminant:
[tex]\[ \sqrt{0.09} = 0.3 \][/tex]
3. Use the quadratic formula to find the roots:
[tex]\[ x = \frac{{-(-0.7) \pm 0.3}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{{0.7 \pm 0.3}}{2} \][/tex]
4. Solve for the two roots:
- For the positive root:
[tex]\[ x = \frac{{0.7 + 0.3}}{2} \][/tex]
[tex]\[ x = \frac{1.0}{2} \][/tex]
[tex]\[ x = 0.5 \][/tex]
- For the negative root:
[tex]\[ x = \frac{{0.7 - 0.3}}{2} \][/tex]
[tex]\[ x = \frac{0.4}{2} \][/tex]
[tex]\[ x = 0.2 \][/tex]
So, the two solutions to the equation [tex]\( x^2 - 0.7x + 0.1 = 0 \)[/tex] are:
[tex]\[ x = 0.2 \quad \text{and} \quad x = 0.5 \][/tex]
Therefore, the solutions are [tex]\( \boxed{0.2 \text{ and } 0.5} \)[/tex].
