If the sum of all the roots of the equation \(e^{2x} - 11e^x - 45e^{-x} + \frac{81}{2} = 0\)
is logeP, then p is equal to _____.
The correct answer is 45
Let \(e^x = t\) then equation reduces to
\(t^2−11t−\frac{45}{t}+\frac{81}{2}=0\)
\(⇒ 2t^3 – 22t^2 + 81t – 45 = 0 …(i)\)
if roots of
\(e^{2x} - 11e^x - 45e^{-x} + \frac{81}{2} = 0\)
are α, β, γ then roots of (i) will be
\(e^{α_1}e^{α_2}e^{α_3} \)
Therefore , by using product of roots
\(e^{α_1+α_2+α_3}=45\)
\(⇒ α_1 + α_2 + α_3 \)
= ln 45
⇒ p = 45
MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K.
MX(s) $\rightleftharpoons M^{+(aq) }+ X^{-}(aq)$; $K_{sp} = 10^{-10}$
If the standard reduction potential for $M^{+}(aq) + e^{-} \rightarrow M(s)$ is $(E^{\circ}_{M^{+}/M}) = 0.79$ V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $E^{\circ}_{X^{-}/MX(s)/M}$ is ____________ mV. (nearest integer)
[Given : $\frac{2.303 RT}{F} = 0.059$ V]
An infinitely long straight wire carrying current $I$ is bent in a planar shape as shown in the diagram. The radius of the circular part is $r$. The magnetic field at the centre $O$ of the circular loop is :

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