On this page, you can find the complete solutions of the second exercise of the Quadratic Equations chapter from Basic Mathematics Grade XIII book published and distributed by Sukunda Pustak Bhawan.
In the above-mentioned book, permutation and combination is the first chapter and has four exercises only. Out of which, this is the solution to the second exercise.
Disclaimer:
Exercise 2
1. Find the equation whose roots are:
a) 3, -2
Solution: Let $\rm \alpha = 3$ and $\rm \beta = -2$ then the required quadratic equation is
$$\rm (x - \alpha )(x - \beta ) = 0$$
$$\rm (x - 3)(x - (-2)) = 0$$
$$\rm x^2 + 2x - 3x - 6 = 0$$
$$\rm \therefore x^2 - x - 6 = 0$$
b) -5, 4
Solution: Let $\rm \alpha = -5$ and $\rm \beta = 4$ then the required quadratic equation is
$$\rm (x - \alpha )(x - \beta ) = 0$$
$$\rm (x - (-5))(x - 4) = 0$$
$$\rm x^2 - 4x + 5x - 20 = 0$$
$$\rm x^2 + x - 20 = 0$$
c) $\rm \sqrt{3}, - \sqrt{3}$
Solution: Let $\rm \alpha = \sqrt{3}$ and $\rm \beta = - \sqrt{3}$ then the required quadratic equation is
$$\rm (x - \alpha )(x - \beta ) = 0$$
$$\rm (x - \sqrt{3})(x - (- \sqrt{3})) = 0$$
$$\rm x^2 + \sqrt{3}x - \sqrt{3} x - \sqrt{3}^2 = 0$$
$$\rm x^2 - 3 = 0$$
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