The equations x^{2}+px+q=0 and x^{2}+qx+p=0(p\neq q) have a common root. What is the value of (p+q) ?

Explanation

Subtract the two equations: (p-q)x + (q-p) = 0 \implies (p-q)(x - 1) = 0. Since p \neq q, the common root is x = 1. Substituting x=1 into the first equation gives 1^2+p(1)+q=0 \implies p+q=-1.

Related questions on Algebra

• If p + q + r = 0, then what is z^{\frac{p^2}{qr}} \times z^{\frac{q^2}{rp}} \times z^{\frac{r^2}{pq}} equal to ?
• What is the value of k for which (k^2 - 5k + 4)x^2 + (k^2 - 3k - 4)x + (k^2 - 4k) = 0 is an identity ?
• If \frac{a^2}{b^2 + c^2} = \frac{b^2}{c^2 + a^2} = \frac{c^2}{a^2 + b^2}, then what is the value of a^4 + b^4 + c^4 equal to ?
• If \frac{1}{x} = \frac{1}{p} + \frac{1}{q}, then what is \frac{pq}{p^2 - q^2}\left(\frac{x + p}{x - p} - \frac{x + q}{x - q}\right) equa...
• If (x - 5) is the HCF of x^2 - x - p and x^2 - qx - 10, then what is the value of (p + q) ?