How do you simplify $sqrt325-sqrt117$?

Let $k+1=x$ So $2(k+1)^3+3(k+1)^2+(k+1)$ =$2x^3+3x^2+x$ =$x(2x^2+3x+1)$ Use {cross product method} It ca be factored as =$x(2x+1)(x+1)$ Back!! =$(k+1)(2k+3)(k+2)$

Given: $x^2+x+2 = 0$ Note that this is in the form: $ax^2+bx+c = 0$ with $a=1$, $b=1$ and $c=2$ This has discriminant $Delta$ given by the...

$4^(2x) *2=315$ $-> 4^(2x) = 315/2$ $->ln(4^(2x)) = ln(315/2)$ $->2xln(4)=ln(157.5)$ $->2x=ln(157.5)/ln(4)$ $-> 2x ~=3.64$ $-> x~=1.82$

1] Take the square root of both sides; 2] consider $164$ as $4xx41$; 3] Evaluate the square root of $4$; Remember that you'll have 2 solutions (equal values but one...

If you are wanting simpler versions of the , then here are a few thoughts... Given a quadratic equation of the form: $ax^2+bx+c = 0$ the roots are...

Assuming you are solving for $v$: $125v^2=t$ $(\cancel(125)v^2)/\cancel(125)=t/125$ $v^2=t/125$ $\sqrt{v^2}=\sqrt{t/125}$ $v=\sqrt{t/(5*5*5)}=\sqrt{t}/\sqrt{25*5}=\sqrt{t}/(5\sqrt{5})$

$sqrt117=sqrt(3^2xx13)=3sqrt13$

$6sqrt(7)-sqrt(4*7)$ $6sqrt(7)-2sqrt(7)=4sqrt(7)$

I don't know if it is much of a simplification but you could rationalize the denominator by multiplying the top and bottom by $sqrt(21)$ $4/sqrt(21) = (4sqrt(21))/(21)$ As I said,...

${(x/y)-(y/x)}/{(1/x)-(1/y)}$ On further solving this problem we get, ${(x^2-y^2)/(xy)}/{(x+y)/(xy)}$ Now dividing both the sides we get, $xy$ as cancelled from both the numerator and denominator. ${(x+y)(x-y)}/(x+y)$ Cancelling $x+y$ from both...