Math, asked by mamta06, 2 months ago

प्रश्न: 3 द्विघात बहुपद x2-53x+6
का शून्यक क्या होगा?

Answers

Answered by hudaattar123
21

Answer:

Answer: बहुपद ax + b, a # 0 का केवल एक शून्यक है, जो उस बिंदु का x-निर्देशांक है, जहाँ y = ax + b का ग्राफ x-अक्ष को प्रतिच्छेद करता है।

Answered by rohitkumargupta
1

Answer:

α= 52.885 and β= 0.115

Step-by-step explanation:

द्विघात बहुपद x2-53x+6 का शून्यक क्या होगा?

Given that a quadratic equation x² - 53x + 6.

To find the zeros of the equation.

So,

    x² - 53x + 6.

Here, a= 1 , b = (-53) and c = 6.

By using the Shreedharya formula,

α = \frac{-b +\sqrt{b^{2}-4ac } }{2a}

=\frac{-(-53)+\sqrt{(-53)^{2}-4(1)(6) } }{2(1)}

=\frac{53+\sqrt{2809-24} }{2}

= \frac{53+\sqrt{2785} }{2}

=\frac{53+52.77}{2}

= \frac{105.77}{2}

= 52.885

\beta  = \frac{-b-\sqrt{b^{2}-4ac } }{2a}

= \frac{-(-53) - \sqrt{(-53)^{2}-4(1)(6) } }{2(1)}

= \frac{53 - \sqrt{2809-24} }{2}

=\frac{53-\sqrt{2785} }{2}

= \frac{53-52.77}{2}

= \frac{0.23}{2}

= 0.115

Therefore the zeros of the given quadratic equation is α= 52.885 and

β= 0.115

THANKS.

#SPJ3

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