Question

if alpha + beta are the zeros of the quadratic polynomial X square + bx + c then evaluate the value of 1 over Alpha minus one over beta

Answer 1

hello friend hope it helps.

Answer 2

Answer:  The answer is $\dfrac{\pm\sqrt{b^2-4ac}}{c}.$

Step-by-step explanation:  Given that $\alpha$ and $\beta$ are the zeroes of the polynomial $ax^2+bx+c.$

We are given to find the value of $\dfrac{1}{\alpha}-\dfrac{1}{\beta}.$

From the relations between roots and coefficients, we have

$\alpha+\beta=-\dfrac{b}{a},\\\\\alpha\times \beta=\dfrac{c}{a}.$

Therefore,

$\left(\dfrac{1}{\alpha}-\dfrac{1}{\beta}\right)^2\\\\\\=\dfrac{(\beta-\alpha)^2}{(\alpha\beta)^2}\\\\\\=\dfrac{(\beta+\alpha)^2-4\alpha\beta}{\frac{c^2}{a^2}}\\\\\\=\dfrac{(-\frac{b}{a})^2-4\times\frac{c}{a}}{\frac{c^2}{a^2}}\\\\\\=\dfrac{b^2-4ac}{c^2}.$

Hence, we have

$\dfrac{1}{\alpha}-\dfrac{1}{\beta}=\dfrac{\pm\sqrt{b^2-4ac}}{c}.$

Thus, completed.