Saved Bookmarks
| 1. |
If alpha and beta are the zeros of the quadratic polynomial x square - 3 x minus 10 then evaluate 1 by Alpha minus 1 by betaplz answer step by step |
|
Answer» Answer: GIVEN that \alphaα and \betaβ are the ZEROES of the polynomial ax^2+bx+c.ax 2 +bx+c. We are given to find the value of \dfrac{1}{\ALPHA}-\dfrac{1}{\beta}. α 1
− β 1
. From the RELATIONS between roots and coefficients, we have \BEGIN{gathered}\alpha+\beta=-\dfrac{b}{a},\\\\\alpha\times \beta=\dfrac{c}{a}.\end{gathered} α+β=− a b
, α×β= a c
.
Therefore, \begin{gathered}(\dfrac{1}{\alpha}-\dfrac{1}{\beta})^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}.\end{gathered} ( α 1
− β 1
) 2
= (αβ) 2
(β−α) 2
= a 2
c 2
(β+α) 2 −4αβ
= a 2
c 2
(− a b
) 2 −4× a c
= c 2
b 2 −4ac
.
Step-by-step explanation: Hence, we have \dfrac{1}{\alpha}-\dfrac{1}{\beta}=\dfrac{\pm\sqrt{b^2-4ac}}{c}. α 1
− β 1
= c ± b 2 −4ac
. Thus, completed |
|