Question

find the sum of the reciprocals of the zeros of the equation X square + PX + q = 0​

Answer 1

Answer:

Let the roots of equation x2+px+q=0 be a and b

We have a+b=−p and ab=q

∴$\frac{a+b}{ab} =\frac{1}{a} +\frac{1}{b} =\frac{-p}{q}$

Step-by-step explanation:

Answer 2

✪ $\huge{\red{\underline{\overline{\mathbf{Question}}}}}$ ✪

→find the sum of the reciprocals of the zeros of the equation X square + PX + q = 0

✪ $\huge{\green{\underline{\overline{\mathbf{Answer}}}}}$ ✪

⇒Given:

• $x^2+px+q=0$

⇒To Find:

• $\frac{1}{\alpha}+\frac{1}{\beta}$

⇒Solution:

⇒$Sum \;of\; Zeroes=\red{\frac{-b}{a}}$

⇒$\alpha + \beta = \frac{-p}{1}$........(1)

And;

⇒$Product\;of \; Zeroes=\red{\frac{c}{a}}$

⇒$\alpha \beta = \frac{q}{1}$........(2)

According to question:-

⇒$\Large{\frac{1}{\alpha}+\frac{1}{\beta}}$

$\leadsto \; \frac{\alpha +\beta}{\alpha \beta}$

Using (1) and (2) here ....

$\leadsto \; \Large{\frac{\frac{-p}{1}}{\frac{q}{1}}}$

$\leadsto \huge{\pink{\overbrace{\underbrace{\purple{-\frac{p}{q}}}}}}$

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Formulas Used :-

• $Product\;of \; Zeroes=\frac{c}{a}$
• $Sum \;of\; Zeroes=\frac{-b}{a}$

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