Answer:
Fourth Option. [tex]x = \frac{5}{4} \pm \frac{\sqrt{41}}{4}[/tex]
Step-by-step explanation:
The given equation is:
[tex]2x^{2}-5x+1=3[/tex]
In order to solve the equation, we need to make the Right hand side equal to Zero. Subtracting 3 from both sides, we get:
[tex]2x^{2}-5x-2=0[/tex]
Since, this is a quadratic equation, we can use the quadratic formula to find the values of x which make the equation true. According to the quadratic formula values of x would be:
[tex]x=\frac{-b \pm \sqrt{b^{2}-4ac} }{2a}[/tex]
Here,
a = Coefficient of Squared term = 2
b = Coefficient of x = -5
c = Constant term = -2
Using these values in the formula, we get:
[tex]x=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(2)(-2)}}{2(2)}\\\\ x=\frac{5 \pm \sqrt{25+16}}{4}\\\\ x=\frac{5 \pm \sqrt{41}}{4}\\\\ x = \frac{5}{4} \pm \frac{\sqrt{41}}{4}[/tex]
Hence, the Fourth Option gives the correct answer for x.
