Question

x² - 6x +4=0 find the root of quadratic formula​

Answer 1

Given quadratic equation:

x² - 6x + 4=0

To Find:

The roots using quadratic formula

Solution:

In x² - 6x + 4=0,

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

Discriminant (D) = b² - 4ac

= (-6)² - 4 x 1 x 4

= 36 - 16

= 20

Now, applying the quadratic formula which is:

$\sf{\longrightarrow\:\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}}$

Substituting the values, we get:

$\sf{\longrightarrow\:\dfrac{6\pm\sqrt{20}}{2}}$

$\sf{\longrightarrow\:\dfrac{6\pm2\sqrt5}{2}}$

$\sf{\longrightarrow\:\dfrac{2(3\pm\sqrt5)}{2}}$

Cancelling the 2s in the numerator and denominator,

$\sf{\longrightarrow\:3\pm\sqrt5}$

The roots are 3 + √5 and 3 - √5.

Final answer:

The roots are 3 + √5 and 3 - √5.

Things to note:

- No real roots will exist if discriminant is less than 0.

- There will be two real and equal roots if the discriminant is equal to 0.

- There will be two distinct and real roots if the discriminant is greater than 0.

Answer 2

$\sf \purple{ \spades \: Given : } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \: \: \: \: \: \: \: \bullet \: \: \: \tt{x}^{2} - 6x + 4 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf \purple{ \spades \: To \: find: } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \: \: \: \: \: \: \: \bullet \: \: \sf Roots \: of \: quadratic \: equation$

$\sf \purple{ \spades \:Solution : } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf \: Quadratic \: Formula$

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

where a = 1 , b = -6 and c = 4.

$\tt \: x = \frac{ \cancel- ( \cancel- 6) \: \pm \sqrt{ ({ - 6})^{2} - 4(1)(4) } }{2(1)} \\ \tt \: x = \frac{ 6 \: \pm \sqrt{ 36 - 16 } }{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\tt \: x = \frac{ 6 \: \pm \sqrt{ 20 } }{2} \\ \tt \:x = \frac{6 \: \pm2 \sqrt{5} }{2} \\ \tt \:x = \frac{ \cancel2(3 \: \pm \sqrt{5}) }{\cancel2} \\ \tt \: x = 3 \: \pm \sqrt{5} \\ \small{ \sf \: Roots \: of \: the \: quadratic \: equation \: are : }\\ \boxed{\tt \:x =3 + \sqrt{5}} \sf \: and \: \boxed{\tt \:x =3 - \sqrt{5}}$