Question

Create a quadratic equation whose roots are -2and1/3

Answer 1

Answer:

Step-by-step explanation:

ax^2+bx+c=0

start putting the value x=-2 and u will get an eqn

then put value x=1/3 and u will get another eqn

the solve these two eqn by subtitution or elimination

and find the value of a and b. the put the value of a and b in eqn ax^2+bx+c=0.

Answer 2

$If\ p(-2) = 0,\ (x + 2)\ should\ be\ a\ factor. \\ \\ If\ p(\frac{1}{3}) = 0,\ (x - \frac{1}{3})\ should\ be\ a\ factor. \\ \\ \\ \therefore\ (x + 2)(x - \frac{1}{3}) \\ \\ = x^2 - \frac{1}{3}x + 2x - \frac{2}{3} \\ \\ = x^2 + \frac{5}{3}x - \frac{2}{3} \\ \\ \\ We\ get\ a\ quadratic\ equation. \\ \\ Multiplying\ the\ polynomial\ with\ multiples\ of\ 3\ to\ avoid\ the\ fractional \\ coefficients\ should\ be\ better.$

$\\ \\ E.\ g.:\ x^2 + \frac{5}{3}x - \frac{2}{3}\ \ multiplied\ with\ 3\ gives\ \ 3x^2 + 5x - 2. \\ \\ 3x^2 + 5x - 2 = (x + 2)(3x - 1) = (x + 2)(x - \frac{1}{3})3 \\ \\ \\ Also,\ x^2 + \frac{5}{3}x - \frac{2}{3}\ \ multiplied\ with\ 6\ gives\ \ 6x^2 + 10x - 4. \\ \\ 6x^2 + 10x - 4 = (x + 2)(6x - 2) = (2x + 4)(3x - 1) = (x + 2)(x - \frac{1}{3})6 \\ \\ And\ so\ on.$

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