Question

3. Find the roots of the following equations:
(i)
x-1/x=3,where x is not equal to zero​

Answer 1

✪AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• A equation
• x - 1/x =3
• x is not equal to 0

${\sf{\green{\underline{\large{To\:Verify}}}}}$

• Root of the equation

${\sf{\pink{\underline{\Large{Explanation}}}}}$

Solving the equation.

=>x - 1/x =3

=>x²-1=3x

=>x²-3x-1

From quadratic formula

$\tt{X=\dfrac{-b\pm{\sqrt{b^2-4ac}}}{2a}}$

$\rightarrow\tt{X=\dfrac{3\pm{\sqrt{9-4\times{1}{-1}}}}{2}}$

$\rightarrow\tt{X=\dfrac{3\pm{\sqrt{9+4}}}{2}}$

$\rightarrow\tt{X=\dfrac{3\pm{\sqrt{13}}}{2}}$

Taking x=+

$\rightarrow\tt{X=\dfrac{3+{\sqrt{13}}}{2}}$

Taking x=-

$\rightarrow\tt{X=\dfrac{3-{\sqrt{13}}}{2}}$

Additional Information

Let the zeroes of the polynomial be$\tt\alpha{and}\beta$

Then,

$\tt\alpha{+}\beta\frac{-b}{a}$

&

$\tt\alpha{\times}\beta{=}\frac{c}{a}$

Here,

a=1

b=-3

C=-1

☞

$\tt\alpha{+}\beta{=}\dfrac{3}{1}$

$\tt\alpha{+}\beta{=}\dfrac{Cofficient\:of\:X}{Cofficient\:of\:x^2}=$

&

$\tt\alpha{\times}\beta{=}\dfrac{-1}{1}$

$\tt{\large\alpha{\times}\beta{=}\dfrac{Constant\:term}{Cofficient\:of\:x^2}}$

Hence,relation verified

Answer 2

Answer:

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Step-by-step explanation:

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