Question

8. The roots of the quadratic equation x +x/1= 3, x not = 0 are​

Answer 1

Correct QuestioN :

The roots of the quadratic equation x + 1/x = 3, x ≠ 0.

To FinD :

The roots of given Equation.

SolutioN :

$\tt : \implies x + \dfrac{1}{x} = 3.$

$\tt : \implies \dfrac{ {x}^{2} + 1}{x} = 3.$

$\tt : \implies {x}^{2} + 1= 3x.$

$\tt : \implies {x}^{2} - 3x + 1= 0.$

☛ Compare With General Equation.

$\tt \dagger \: \: \: \: \: a{x}^{2} + bx + c= 0.$

Where as,

• a = 1.
• b = - 3.
• c = 1.

☞ Let Find the value of x by Quadratic Formula.

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

$\tt : \implies x = \dfrac{ 3\pm \sqrt{ {( - 3)}^{2} - 4 \times 1 \times 1} }{2}$

$\tt : \implies x = \dfrac{ 3\pm \sqrt{ 9 - 4 } }{2}$

$\tt : \implies x = \dfrac{ 3\pm \sqrt{ 5} }{2}$

$\rule{90}3$

★ Either,

$\tt : \implies x = \dfrac{ 3+ \sqrt{ 5} }{2}$

★ Or,

$\tt : \implies x = \dfrac{ 3\pm -{ 5} }{2}$

Therefore, the roots of given Equation is 3 ± √5 /2.

Answer 2

Answer:

Step-by-step explanation:

ANSWER:

First we will solve the equation and make it into Quadratic Equation.

$x+\frac{1}{x}=3\\\frac{x^2+1}{x}=3$

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

Now we know the Quadratic Formula as:

$x = \frac{-b\pm\sqrt{D}}{2a}$

Where D is

$D = b^2-4ac$

Calculating D first,

a = 1, b = -3 , c = 1

$D = (-3)^2-4(1)(1)$

D = 9 - 4

D = 5

Placing D in Quadratic Formula,

$\implies\frac{-(-3)\pm\sqrt{5}}{2(1)}$

$\frac{3+\sqrt{5}}{2}$ and $\frac{3-\sqrt{5}}{2}$ are the zeroes of polynomial.

Things to Remember:-

$x = \frac{-b\pm\sqrt{D}}{2a}$ is the Quadratic Formula used.

$D = b^2-4ac$ is used to find the Discriminant.