Let α and β be the roots of (p2 + q2) x2 – 2q(p + r)x + q2 + r2 = 0
∴ α + β> 0 and αβ> 0
Also, it has a common root with x2 + 2x – 8 = 0
∴ The common root between above two equations is 4.
16(p2 + q2) – 8q(p + r) + q2 + r2 = 0
(16p2 – 8pq + q2) + (16q2 – 8qr + r2) = 0
(4p – q)2 + (4q – r)2 = 0
q = 4p and r = 16p
\(∴\frac{q^2+r^2}{p^2}=\frac{16p^2+256p^2}{p^2}=272\)
So, the correct is 272.
If the roots of the quadratic equation \( ax^2 + bx + c = 0 \) are real and equal, then:
A polynomial that has two roots or is of degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b, and c are the real numbers.
Consider the following equation ax²+bx+c=0, where a≠0 and a, b, and c are real coefficients.
The solution of a quadratic equation can be found using the formula, x=((-b±√(b²-4ac))/2a)
Read More: Nature of Roots of Quadratic Equation