What Is The Solution To Square Root 3x - 2 Equals 5

What is the solution to (x^3 -3x^2 + 2x - 1/ x^2 - 4x + 4)dx?

What is the solution to (x^3–3x^2 + 2x - 1 / x^2 - 4x + 4)dx?Right, so you want the (indefinite) integral of an expression with respect to x.Problem: What is your expression? In particular, what is being divided by what?As written, your expression is actually: [math]x^3 - 3x^2 + 2x - \frac {1}{x^2} - 4x + 4[/math]Given that there are two terms in [math]x[/math], I suspect this is not what you meant. So, did you mean:x^3–3x^2 + 2x - 1 / (x^2 - 4x + 4) or(x^3–3x^2 + 2x - 1) / (x^2 - 4x + 4)?I will assume the latter.[math]I = \int \frac {x^3 - 3x^2 + 2x - 1}{x^2 - 4x + 4} dx[/math]Factorising the numerator, we have:[math]I = \int \frac {(x^2 - 4x + 4)(x + 1) + 2x - 5}{x^2 - 4x + 4} dx[/math][math]= \int \left( x + 1 + \frac {2x - 4 - 1}{x^2 - 4x + 4} \right) dx[/math]Factorising the denominator and splitting the fraction into two parts, we have:[math]I = \int \left( x + 1 + \frac {2x - 4}{(x - 2)^2} - \frac {1}{(x - 2)^2} \right) dx[/math]Note that the numerator of the first fraction is the derivative of the denominator;[math]\int \frac {f'(x)}{f(x)} = ln[f(x)][/math], thus[math]I = \frac {1}{2}x^2 + x + ln[(x - 2)^2] + \frac {1}{x - 2} + c[/math]where c is the constant of integration[math]= \frac {x(x + 2) + 4ln|x - 2|}{2} + \frac {1}{x - 2} + c[/math]

If the roots of the equation 3x^2+9x+2 are in ratio m:n then find (m/n) ^1/2+(n/m) ^1/2?

The answer is not [math]-\sqrt{\frac{3}{2}}[/math], as a quick glance might indicate. The expression to be evaluated is the sum of two square roots—it cannot be negative. It has to be positive (or zero).If the zeros of the quadratic polynomial[math]3x^2+9x+2[/math]are in the ratio [math]m:n[/math], assume that the roots are [math]km[/math] and [math]kn[/math]. Assume that [math]m[/math] and [math]n[/math] are positive.[math]m,n>0\implies k<0[/math]The fact that [math]k[/math] is negative follows from the observation that both zeros of the quadratic polynomial are negative. Now, manipulate the expression to be evaluated.[math]\sqrt{\cfrac{m}{n}}+\sqrt{\cfrac{n}{m}}=\cfrac{m+n}{\sqrt{mn}}[/math][math]\implies\sqrt{\cfrac{m}{n}}+\sqrt{\cfrac{n}{m}}=\cfrac{m+n}{\sqrt{mn}}\times\cfrac{k}{k}[/math][math]\implies\sqrt{\cfrac{m}{n}}+\sqrt{\cfrac{n}{m}}=\cfrac{k(m+n)}{k\sqrt{mn}}[/math]Here comes the tricky part. Square roots are, by definition, positive. Hence, to send [math]k[/math] inside the radical, we must introduce negation somehow.[math]k<0[/math][math]\implies k=-\sqrt{k^2}[/math][math]\implies\sqrt{\cfrac{m}{n}}+\sqrt{\cfrac{n}{m}}=\cfrac{k(m+n)}{-\sqrt{k^2}\sqrt{mn}}[/math][math]\implies\sqrt{\cfrac{m}{n}}+\sqrt{\cfrac{n}{m}}=-\cfrac{k(m+n)}{\sqrt{k^2mn}}[/math][math]\implies\sqrt{\cfrac{m}{n}}+\sqrt{\cfrac{n}{m}}=-\cfrac{km+kn}{\sqrt{km\times kn}}[/math]The sum and product of [math]km[/math] and [math]kn[/math] can be found easily, because they are the zeros of[math]3x^2+9x+2[/math]the given quadratic expression.[math]km+kn=-\cfrac{9}{3}=-3[/math][math]km\times kn=\cfrac{2}{3}[/math][math]\implies\sqrt{\cfrac{m}{n}}+\sqrt{\cfrac{n}{m}}=-\cfrac{-3}{\sqrt{\frac{2}{3}}}[/math][math]\implies\sqrt{\cfrac{m}{n}}+\sqrt{\cfrac{n}{m}}=3\sqrt{\cfrac{3}{2}}[/math]You’d have got the same answer if you had assumed that [math]m,n<0[/math] and [math]k>0[/math].