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The roots of the equation $x^{2} - 6 x + 4 = 0$ are [[1]] and [[2]].

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answer is 3 - 5, 3 + 5.

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Detailed Solution

The roots of the equation

x^{2} - 6 x + 4 = 0

are

3 - \sqrt{5}

and

3 - \sqrt{5}

.
Given quadratic equation is,

x^{2} - 6 x + 4 = 0

.
We know that,
Discriminant for the quadratic equation

a x^{2} + bx + c = 0, a \neq 0

where a, b and c are real numbers is given by,

D = b^{2} - 4 ac

.
On comparing the given equation

x^{2} - 6 x + 4 = 0

with standard form we get,

a = 1, b = - 6, c = 4

Then,

D = b^{2} - 4 ac

⟹ D = {(- 6)}^{2} - 4 \times 1 \times 4

⟹ D = 36 - 16

⟹ D = 20 > 0

Discriminant is greater than zero which means the roots of the equation exist.
Since, quadratic formula for the quadratic equation

a x^{2} + bx + c = 0, a \neq 0

where a, b and c are real numbers is given by,

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

Then,

x = \frac{- (- 6) \pm \sqrt{20}}{2 \times 1}

⟹ x = \frac{6 \pm 2 \sqrt{5}}{2}

⟹ x = 3 \pm \sqrt{5}

On taking positive sign,

⟹ x = 3 + \sqrt{5}

On taking negative sign,

⟹ x = 3 - \sqrt{5}

Therefore, the roots of the equation

x^{2} - 6 x + 4 = 0

are

3 + \sqrt{5}

and

3 - \sqrt{5}

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