Question

Solution of the quadratic ײ+×+1=0 is
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Answer 1

Required Answer:-

Given:

• x² + x + 1 = 0

To find:

• The values of x.

Solution:

Given that,

$\rm {x}^{2} + x + 1 = 0$

Here,

$\rm \implies a = 1$

$\rm \implies b= 1$

$\rm \implies c= 1$

Where,

a = coefficient of x²

b = coefficient of x.

c = coefficient of x^0

So,

$\rm x _{1,2} = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a}$

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

$\rm \implies x = \dfrac{ - 1 \pm \sqrt{1 - 4} }{2}$

$\rm \implies x = \dfrac{ - 1 \pm \sqrt{ - 3} }{2}$

$\rm \implies x = \dfrac{ - 1 \pm i\sqrt{3} }{2}$

So,

$\rm \implies x_{1} = \dfrac{ - 1 + i\sqrt{3} }{2}$

And,

$\rm \implies x_{2} = \dfrac{ - 1 - i\sqrt{3} }{2}$

Hence, the roots of the quadratic equation are (-1 + i√3)/2 and (-1 - i√3)/2

Note:

• i = √-1

Quadratic Formula:

$\rm x _{1,2} = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a}$

where, Discriminant = b² - 4ac.

Answer 2

$\sf{x^2+x+1=0}$

$\implies\sf{\displaystyle x_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\cdot \:1\cdot \:1}}{2\cdot \:1}}$

$\implies\sf{\displaystyle x_{1,\:2}=\frac{-1\pm \sqrt{3}i}{2\cdot \:1}}$

$\implies\sf{\displaystyle x_1=\frac{-1+\sqrt{3}i}{2\cdot \:1},\:x_2=\frac{-1-\sqrt{3}i}{2\cdot \:1}}$

$\boxed{\implies\sf{\displaystyle x=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}}}$