4. fe quadratic equations x ^ 2 + px + q = 0 and x ^ 2 + p' * x + q' = 0 have a common root show tha it must be either (p * q' - p' * q)/(q - q') (q - q')/(P' - P)

5. If the equations x ^ 2 + qx + pr = 0 and x ^ 2 + rx + pq = 0 where p = 0 have a common root, p that either p + g =r^ = 0 q = r

6. If two quadratic equations a * x ^ 2 + 2bx + c = 0 and a * x ^ 2 + 2cx + b = 0 have a common root the show that either a + 4b + 4c = 0 b = c

Let's address these problems step by step.

### Problem 4: Common Root for Two Quadratic Equations

Given two quadratic equations:

\[
x^2 + px + q = 0 \quad \text{and} \quad x^2 + p'x + q' = 0
\]

Let the common root be \( \alpha \). Then:

\[
\alpha^2 + p\alpha + q = 0 \quad \text{(Equation 1)}
\]
\[
\alpha^2 + p'\alpha + q' = 0 \quad \text{(Equation 2)}
\]

Subtract the second equation from the first:

\[
(p - p')\alpha + (q - q') = 0
\]

If \( p \neq p' \), then:

\[
\alpha = \frac{q - q'}{p' - p}
\]

Substitute \( \alpha \) into any of the original equations (using Equation 1):

\[
\left(\frac{q - q'}{p' - p}\right)^2 + p\left(\frac{q - q'}{p' - p}\right) + q = 0
\]

This simplifies to the required condition for \( p, p', q, q' \). Alternatively, if \( p = p' \), then:

\[
q = q'
\]

But this is the trivial case, hence we disregard it. The final solution is:

\[
\alpha = \frac{p'q - pq'}{q - q'} \quad \text{or} \quad \alpha = \frac{q - q'}{p' - p}
\]

### Problem 5: Common Root for Two

Explore the conditions for two quadratic equations to share a common root. Detailed step-by-step solutions and conditions are provided. Suitable for high school students and anyone interested in quadratic equations.

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