is m the slope? what formula i may use to solve for this?
Just because the textbook says $\displaystyle y = mx + b$ does not mean you should automatically assume $\displaystyle m$ is the slope. Besides, what slopes are mentioned in this problem?
no, im solving it this way that i have to divide both sides of the equation by x^2+ 1 then m = 1 - 4 sq. root of 3 x underneath x^2+1 sir may i know the next move for the value of m?
thanks
Okay, so $\displaystyle m = \frac{1 - 4 \sqrt{3}x}{x^2 + 1}$. But does get you any closer to a solution?
I would start with the original equation and expand the LHS, yielding
$\displaystyle mx^2 + m = 1 - 4 \sqrt{3} x$
$\displaystyle mx^2 + 4 \sqrt{3} x + (m-1) = 0$
This has two equal roots if and only if the discriminant is zero.
sir im solving for the discriminant.... that a = m, b = 4, and c = (m-1) is this correct? the equate to zero.
D = 4^2 - 4(m)(m-1)
D = 16 - 4m^2 + 4m
0 = 16 - 4m^2 + 4m
0 = -4(m^2-m-4)
0 = m^2-m-4 sir this not a factorable expression... what is the next sir?
thanks
There should be a $\displaystyle 4 \sqrt{3}$ in there, i.e.
$\displaystyle D = (4 \sqrt{3})^2 - 4m(m-1) = 0$
$\displaystyle 48 - 4m^2 + 4m = 0$
$\displaystyle m^2 - m - 12 = 0$, this factors. If it didn't factor, how would you solve it then?
if i were to solve it sir , i will use quadratic formula. is it ok?
factor like ( m - 4) (m + 3) = 0
then m = 4 or m = - 3
is that right sir?
in that case, check your answer:
m = 4:
4(x^{2} + 1) = 1 - (4√3)x
4x^{2} + 4 = 1 - (4√3)x
4x^{2} + (4√3)x + 3 = 0
since we are supposed to have 2 equal roots, this should factor as:
(2x + √3)^{2} = 0. does it?
m = -3:
-3(x^{2} + 1) = 1 - (4√3)x
-3x^{2} - 3 = 1 - (4√3)x
3x^{2} + (4√3)x + 4 = 0
again, this "ought" to factor as:
(√3x + 2)^{2} = 0. does it?