Maths - if.p r are rational numbers and p not equal r ,than roots of the equation{p2 -q2} x2-?

Does the x2 mean x²? If so and the equation you want to solve is:

(p²-q²)x² - (q²-r²)x + (r²-p²) = 0

...then this is a standard quadratic and you can solve it with the formula:

x = (-b ± √ (b² - 4ac)) / 2a where:

a = p²-q²
b = q²-r²
c = r²-p²

Replacing a, b and c with your coefficients:

x = (-(q²-r²) ± √ ( (q²-r²)² - 4(p²-q²)(r²-p²) )) / 2(p²-q²)

Expanding the brackets:

x = (r² - q² ± √ (q^4 - 2qr² + r^4 - 4pr² + 4p^4 + 4qr² - 4qp² )) / 2p² - 2q²

We've got two terms in qr² so simplifying those:

x = (r² - q² ± √ (q^4 + r^4 + 4p^4 + 2qr² - 4qp² - 4pr²)) / 2p² - 2q²

Trying to simplify the bit after the √ into two terms multiplied by each other, we can see the polarity of the coefficients of q and r is the same (because multiplying them gives +2qr²), the polarity of q and p is different and p and r are different, therefore q and r are + and p is -, or q and r are - and p is +. Looking at the values of the coefficients, q and r are 1 or -1 because there's a single q^4 and r^4 and p is 2 or -2 because of 4p^4. Trying out the options:

(2p² - r² - q²)² expands to:

4p^4 - 2pr² - 2pq² - 2pr² + r^4 + qr² - 2pq² + qr² + q^4

Which can be rewritten as:

q^4 + r^4 + 4p^4 + 2qr² - 4gp² - 4pr²

Which is what you have in your brackets so we have:

x = (r² - q² ± √ (2p² - r² - q²)²) / 2p² - 2q²

We can now remove the square root because it's the square root of something squared:

x = (r² - q² ± (2p² - r² - q²)) / 2p² - 2q²

Or:

x1 = (r² - q² + (2p² - r² - q²)) / 2p² - 2q²
x2 = (r² - q² - (2p² - r² - q²)) / 2p² - 2q²

x1 = (2p² - 2q²) / (2p² - 2q²)
x2 = (2r² - 2p²) / (2p² - 2q²)

Removing the twos.

x1 = (p² - q²) / (p² - q²)
x2 = (r² - p²) / (p² - q²)