Question

x² - 4x + 13 = 0 ,
solve this equation​

Answer 1

x² - 4x + 13 = 0

______________ [ GIVEN ]

• We have to solve this equation and find the value of x.

___________________________

x² - 4x + 13 = 0

Now, the above equation is in the ax² + bx + c.

So here, a = 1, b = -4 and c = 13

We know that

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

Put the known values in above formula

→ x = $\frac{ - (-4) \: \pm \: \sqrt{ {(-4)}^{2} \: - \: 4(1)(13)} }{2(1)}$

→ x = $\frac{ 4 \: \pm \: \sqrt{ 16 \: - \: 52} }{2}$

→ x = $\frac{ 4 \: \pm \: \sqrt{-\:36} }{2}$

Now, in square root the value which came is negative.

It means $\textbf{no real roots exist.}$

______________________________

OR

x² - 4x + 13 = 0

______________ [ GIVEN ]

• We have to solve this equation and find the value of x.

___________________________

x² - 4x + 13 = 0

Now, the above equation is in the ax² + bx + c.

So here, a = 1, b = -4 and c = 13

We know that

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

Put the known values in above formula

→ x = $\frac{ - (-4) \: \pm \: \sqrt{ {(-4)}^{2} \: - \: 4(1)(13)} }{2(1)}$

→ x = $\frac{ 4 \: \pm \: \sqrt{ 16 \: - \: 52} }{2}$

→ x = $\frac{ 4 \: \pm \: \sqrt{-\:36} }{2}$

We know that √36 = 6. (as 6 × 6 = 36) But here the value of square root is in negative. So, we take "i". So, that it's negative sign can be removed. Here , i = iota.

→ x = $\frac{ 4 \: \pm\: 6i}{2}$

→ x = $\frac{ 4 \: + 6i}{2}$

Take 2 common from it

→ x = $\frac{ 2(2 \: + 3i)}{2}$

→ x = 2 + 3i

Similarly,

→ x = $\frac{ 4 \: - 6i}{2}$

→ x = 2 - 3i

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x = 2 ± 3i

_________ [ ANSWER ]

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Answer 2

Hi there!

Ist Method :-

→x² - 4x +13 =0

→x² - 4x + 4 + 9 =0

We know that (a-b)² = a²+b²-2ab

→(x - 2 )² + 9 = 0

→(x-2)² - 9i² = 0 ( °•° i²= -1)

→(x - 2)² - (3i)² = 0

→{(x -2) - 3i} {(x-2)+3i} = 0

→(x-2-3i) (x - 2+3i) =0

→x - 2 - 3i = 0 or x - 2 + 3i = 0

→x = 2 + 3i or x = 2 - 3i

Hence, the roots are 2+3i and 2-3i

Second method:-

Refer to that attachment ↑

Thankyou :)