how to find a quadratic equation is standard from whoes rooot are given by

r₁=3+√2 and r₂=3-[FONT='Arial','sans-serif']√2

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- Jun 10th 2008, 05:43 PMtejashtajneed help fotr math
- Jun 10th 2008, 06:15 PMtopsquark
You know the values of the two roots, and we know that one form of the quadratic for this will be given by

$\displaystyle (x - r_1)(x - r_2)$

Thus

$\displaystyle (x - (3 + \sqrt{2} ))(x - (3 - \sqrt{2}))$

The simplest way to expand this (in my opinion) is to write this as

$\displaystyle ((x - 3) - \sqrt{2} ) ((x - 3) + \sqrt{2})$

which is in the form of $\displaystyle (a - b)(a + b) = a^2 - b^2$,

$\displaystyle ((x - 3) - \sqrt{2} ) ((x - 3) + \sqrt{2}) = (x - 3)^2 - (\sqrt{2})^2$

$\displaystyle = x^2 - 6x + 9 - 2 = x^2 - 6x + 7$

as you can check.

-Dan - Jun 10th 2008, 06:40 PMtejashtaj
write a quadratic of the form f(x)=a[FONT='Arial','sans-serif']x² +bx +c for the quadratic function whose graph passes through the point (o,8),(8,0) and (-2,5).[/FONT]

[FONT='Arial','sans-serif'](Hint find the values of a,b and c!)[/FONT] - Jun 10th 2008, 08:04 PMReckoner
Substitute the given values into the general equation, and you will end up with a system of linear equations in $\displaystyle a,\;b,\text{ and }c$:

$\displaystyle \begin{array}{rcl}

(x,\;y) &\rightarrow& y = ax^2 + bx + c\\

(0,\;8) &\rightarrow& 8 = a(0)^2 + b(0) + c\\

(8,\;0) &\rightarrow& 0 = a(8)^2 + b(8) + c\\

(-2,\;5) &\rightarrow& 5 = a(-2)^2 + b(-2) + c

\end{array}$

and you get the following system:

$\displaystyle \left\{\begin{array}{rcrcrcr}

64a & + & 8b & + & c & = & 0\\

4a & - & 2b & + & c & = & 5\\

&&&& c & = & 8

\end{array}\right.$

Now solve for the needed coefficients.