The condition that one root of the equation $ax^{2}+bx+c=0$ exceeds the other by ${p}$ is
- ${a}^{2}{p}^{2}=4{a}{c}$
- ${a}^{2}{p}^{2}={b}^{2}+4{a}{c}$
- ${a}^{2}{p}^{2}={b}^{2} -4ac$
- $ ap ={b}^{2}+4{a}{c}$
Let roots be alpha and alpha + p. Sum of roots = -b/a = 2*alpha + p, so alpha = (-b/a - p)/2. Product of roots = c/a = alpha(alpha + p). Substituting alpha: c/a = ((-b/a - p)/2) * ((-b/a + p)/2) = (b^2/a^2 - p^2) / 4. 4c/a = b^2/a^2 - p^2. Multiplying by a^2: 4ac = b^2 - a^2*p^2. Rearranging gives a^2*p^2 = b^2 - 4ac.
Let the roots of the quadratic equation be alpha and beta, and their difference p equals alpha minus beta. Squaring this difference gives p squared equals (alpha plus beta) squared minus 4 times alpha times beta. Using the sum and product formulas for roots, we substitute alpha plus beta equals -b/a and alpha beta equals c/a. This yields p squared equals (b squared - 4ac) divided by a squared, which rearranges to a squared p squared equals b squared - 4ac.