a,b,c belong to R and a is not equal to 0. if equations ax2+bx+c=0 anf 2x2-5x+8+0 have common roots find value of a+c/-2b

To solve this problem, we need to determine the common roots of the quadratic equations:

1. \(ax^2 + bx + c = 0\)
2. \(2x^2 - 5x + 8 = 0\)

and then find the value of \(\frac{a+c}{-2b}\).

### Step-by-Step Solution

#### Step 1: Find the roots of the second equation

Consider the equation:

\[
2x^2 - 5x + 8 = 0
\]

To find the roots, use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \(a = 2\), \(b = -5\), and \(c = 8\).

Substitute these values into the formula:

\[
x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 8}}{2 \cdot 2}
\]

Simplify:

\[
x = \frac{5 \pm \sqrt{25 - 64}}{4}
\]

\[
x = \frac{5 \pm \sqrt{-39}}{4}
\]

Since the discriminant (\(b^2 - 4ac = -39\)) is negative, the roots are complex. Let's write them in the form:

\[
x = \frac{5 \pm i\sqrt{39}}{4}
\]

The roots are:

\[
x_1 = \frac{5 + i\sqrt{39}}{4}, \quad x_2 = \frac{5 - i\sqrt{39}}{4}
\]

#### Step 2: Use the common roots in the first equation

Since the roots \(x_1\) and \(x_2\) are common to both equations, \(x_1\) and \(x_2\) must also satisfy the equation:

\[
ax^2 + bx + c = 0
\]

For a quadratic with complex roots, if the roots \(p\) and \(q\) are common, and given by:

\[
p = \frac{5 + i\sqrt{39}}{4}, \quad q = \frac{5 - i\sqrt{39}}{4}
\]

The sum and product of the roots for the second equation are:

- Sum of roots, \(S\): \(p + q = \frac{5 + i\sqrt{39}}{4} + \frac{5 - i\sqrt{39}}{4} = \frac{10}{4} = \frac{5}{2}\)
- Product of roots, \(P\): \(p \cdot q = \left(\frac{5 + i\sqrt{39}}{4}\right) \left(\frac{5 - i\sqrt{39}}{4}\right) = \frac{25 - (-39)}{16} = \frac{64}{16} = 4\)

For the equation \(ax^2 + bx + c = 0\), using Vieta's formulas:

- Sum of roots: \(-\frac{b}{a} = \frac{5}{2}\)
- Product of roots: \(\frac{c}{a} = 4\)

#### Step 3: Solve for \(a\), \(b\), and \(c\)

From \(-\frac{b}{a} = \frac{5}{2}\):

\[
b = -\frac{5a}{2}
\]

From \(\frac{c}{a} = 4\):

\[
c = 4a
\]

#### Step 4: Calculate \(\frac{a+c}{-2b}\)

Now substitute the values of \(b\) and \(c\):

\[
\frac{a + c}{-2b} = \frac{a + 4a}{-2(-\frac{5a}{2})}
\]

Simplify the expression:

\[
\frac{5a}{5a} = 1
\]

#### Final Answer

\[
\frac{a+c}{-2b} = 1
\]

So, the value of \(\frac{a+c}{-2b}\) is \(\boxed{1}\).

Would you like to see more details or have any questions?

### Related Questions

1. How do you determine if a quadratic equation has real or complex roots?
2. How does the quadratic formula derive from the standard form of a quadratic equation?
3. Can a quadratic equation have more than two roots? Why or why not?
4. What is the significance of the discriminant in a quadratic equation?
5. How do Vieta's formulas relate to the coefficients of a quadratic equation?

### Tip

Always check the discriminant (\(b^2 - 4ac\)) of a quadratic equation to determine whether the roots are real or complex.

Explore how to find common roots of quadratic equations ax^2 + bx + c = 0 and 2x^2 - 5x + 8 = 0 using Vieta's formulas and the quadratic formula. Learn step-by-step with examples and calculations. Suitable for advanced high school students.

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