# How to solve algebraic surds by factorising?

• Feb 6th 2009, 05:07 PM
positiveion
How to solve algebraic surds by factorising?
This is the answer in the book but I don't know how they got from

x^2-5root(2)X+12=0

to

(x-3root(2))(x-2root(2))

Thank you!
• Feb 6th 2009, 06:02 PM
elliotyang
well, you can use the quadratic formula to find the roots of the equation, which the roots are
[-b+sqrt(b^2-4ac)]/2a and [-b-sqrt(b^2-4ac)]/2a

where a=1, b=-5root(2), c=12

after you get the roots, rewrite your equation in the form of
(x-root1)(x-root2)=0
• Feb 7th 2009, 05:16 AM
HallsofIvy
First note that $(x+x_0)(x+x_1)= x^2+ (x_0+ x_1)x+ x_0x_1$. That means that in order to factor a polynomial of the form $x^2+ ax+ b$ you look for numbers $x_0$ and $x_1$ such that $x_0+ x_1= a$ and $x_0x_1= b$. In this particular example $a= -5\sqrt{2}$ and $b= 12$ so you need to solve $x_0+ x_1= -5\sqrt{2}$ and $x_0x_1= 12$. Since that product does NOT have a $\sqrt{2}$ in it, it is clear that both $x_0$ and $x_1$ must have a factor of $\sqrt{2}$.

So, now, we are thinking $x_0= m\sqrt{2}$ and $x_1= n\sqrt{2}$ for some numbers m and n. $x_0+ x_1= (m+n)\sqrt{2}= -5\sqrt{2}$ and $x_0x_1= 2mn= 12$ so we now have m+ n= 5 and mn= 6. -2 and -3 satisfy those equations so we can take $x_0= -2\sqrt{2}$ and $x_1= -3\sqrt{2}$ and have $x^2- 5\sqrt{2}x+ 12= (x- 2\sqrt{2})(x- 3\sqrt{2})$