Question

solve Using the quadratic formula:x^2_4x+1=0​

Answer 1

Answer:

Given Equation:

→ x² - 4x + 1 = 0

General Equation of a Quadratic Polynomial:

→ ax² + bx + c

Comparing both the equations we get:

⇒ a = 1, b = -4, c = 1

The quadratic formula in terms of a, b and c is given as:

$\boxed{ \bf{x = \dfrac{ - b \pm \sqrt{b^2 - 4ac}}{2a}}}$

Substituting the given information we get:

$x = \dfrac{ -(-4) \pm \sqrt{16 - 4(1)(1)}}{2\times1}\\\\\\x = \dfrac{ 4 \pm \sqrt{12}}{2}\\\\\\x = \dfrac{ 4 \pm 2\sqrt{3}}{2}\\\\\\\boxed{ \bf{x = 2 \pm \sqrt{3}}}$

Hence the values of 'x' or roots of the given quadratic equation are:

→ x = 2 + √3

→ x = 2 - √3

Answer 2

Solution:

$x = 2 - \sqrt{3 \: } \: and \: x = 2 + \sqrt{3 \: }$

Step-by-step explanation:

The quadratic formula states:

For ax² + bx + c = 0, the values of x which are the solutions to the equation are given by:

$x = \frac { - b \: ±\sqrt{ {b}^{2} - (4ac) \: } }{2 \: . \: a} \: \:$

Substituting:

1 for a

-4 for b

1 for c gives:

$x = \frac{ - ( - 4) ± \sqrt{ {( - 4) }^{2} - (4 \: . \: 1 \: . \: 1)} }{2 \: . \: 1}$

$x = \frac{ \: \: 4± \sqrt{16 - 4 \: \: } }{2}$

$x = \frac{ \: 4± \sqrt{12 \: } }{2}$

$x = \frac{ \: 4 - \sqrt{4 \: . \: 3 \: }}{2} \: and \: x = \frac{ \: 4 \: + \: \sqrt{4 \: . \: 3 \: }}{2} \:$

$x = \frac{ \: 4 - \sqrt{4 }\sqrt{3} \: }{2} \: and \: x = \frac{ \: 4 \: + \: \sqrt{4 }\sqrt{3} \: }{2}$

$x = \frac{ \: 4 - 2 \sqrt{3} \: }{2} \: and \: x = \frac{ \: 4 + 2 \sqrt{3} \: }{2}$

$x = \frac{4}{2} - \frac{ \: 2 \sqrt{3 \: } }{2} \: and \: x = \frac{4}{2} - \frac{ \: 2 \sqrt{3 \: } }{2}$

$x = 2 - \sqrt{3 \: } \: and \: x = 2 + \sqrt{3 \: }$