Math, asked by rekhaceig, 7 months ago

2x^2-3x+13
Factorized answer pls​

Answers

Answered by udayagrawal49
0

Answer:

The factors of f(x) are [x - (\frac{3+i\sqrt{95}}{4})] and [x - (\frac{3-i\sqrt{95}}{4})].

Step-by-step explanation:

Let f(x) = 2x²-3x+13 = 0.

w.k.t., D = b²-4ac

⇒ D = 3²-4×2×13 = 9-104 = -95 < 0

⇒ f(x) has imaginary roots.

w.k.t., x = (-b ± √D)/2a

⇒ x = [-(-3) ± √(-95)] / 2×2

⇒ x = [3 ± i√95] / 4

⇒ Factors of f(x) are [x - (\frac{3+i\sqrt{95}}{4})] and [x - (\frac{3-i\sqrt{95}}{4})].

Please mark it as Brainliest.

Answered by gouravgupta65
1

Answer:

The factors of f(x) are [x - (\frac{3+i\sqrt{95}}{4})][x−(

4

3+i

95

)] and [x - (\frac{3-i\sqrt{95}}{4})][x−(

4

3−i

95

)] .

Step-by-step explanation:

Let f(x) = 2x²-3x+13 = 0.

w.k.t., D = b²-4ac

⇒ D = 3²-4×2×13 = 9-104 = -95 < 0

⇒ f(x) has imaginary roots.

w.k.t., x = (-b ± √D)/2a

⇒ x = [-(-3) ± √(-95)] / 2×2

⇒ x = [3 ± i√95] / 4

⇒ Factors of f(x) are [x - (\frac{3+i\sqrt{95}}{4})][x−(

4

3+i

95

)] and [x - (\frac{3-i\sqrt{95}}{4})][x−(

4

3−i

95

)] .

Please mark it as Brainliest.

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