Question

एक द्विघात बहुपद ज्ञात कीजिए जिसके शून्यक बहुपद X²-X-1के शुन्यकों का वर्ग है।​

Answer 1

Answer:

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Step-by-step explanation:

Answer 2

The equation of polynomial=$x^2-3x+1=0$

Step-by-step explanation:

Given polynomial

$x^2-x-1$

Quadratic formula:$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Where a=Coefficient of $x^2$

b=Coefficient of x

c=Constant term

a=1,b=-1 and c=-1

Using the formula

$x=\frac{1\pm\sqrt{1-4(1)(-1)}}{2}$

$x=\frac{1\pm\sqrt 5}{2}$

The roots of the given polynomial are

$x=\frac{1+\sqrt 5}{2}, x=\frac{1-\sqrt 5}{2}$

$\alpha=x^2=(\frac{1+\sqrt 5}{2})^2=\frac{1+5+2\sqrt 5}{4}=\frac{6+2\sqrt 5}{4}=\frac{3+\sqrt 5}{2}$

$\beta=(\frac{1-\sqrt 5}{2})^2=\frac{1+5-2\sqrt 5}{4}=\frac{6-2\sqrt 5}{4}=\frac{3-\sqrt 5}{2}$

By using formula

$(a+b)^2=a^2+b^2+2ab$

$(a-b)^2=a^2+b^2-2ab$

$\alpha+\beta=\frac{3+\sqrt 5}{2}+\frac{3-\sqrt 5}{2}$

$\alpha+\beta=\frac{3+3+\sqrt5 -\sqrt 5}{2}=\frac{6}{2}=3$

$\alpha\beta=\frac{3+\sqrt 5}{2}\times\frac{3-\sqrt 5}{2}=\frac{3^2-(\sqrt 5)^2}{4}=\frac{9-5}{4}=\frac{4}{4}=1$

By using the formula

$(a+b)(a-b)=a^2-b^2$

General equation of quadratic polynomial

$x^2-(\alpha+\beta)+\alpha\beta$

Substitute the values

$x^2-3x+1=0$

This is required equation of polynomial.

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