Question

factorise it 2x^2-7x+1

Answer 1

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Answer 2

Answer:

The roots are $x=\frac{7+\sqrt{41}}{4},\frac{7-\sqrt{41}}{4}$

Step-by-step explanation:

Given : Quadratic equation -$2x^2-7x+1$

To factorize the quadratic equation :

Solution :

Solving by discriminant method  

General form - $ax^2+bx+c=0$

$D=b^2-4ac$

Solution is $x=\frac{-b\pm\sqrt{D}}{2a}$

Equation is $2x^2-7x+1$

where a=2 , b=-7, c=1

$D=b^2-4ac$

$D=(-7)^2-4(2)(1)$

$D=49-8$

$D=41$

Solution is $x=\frac{-b\pm\sqrt{D}}{2a}$

$x=\frac{-(-7)\pm\sqrt{41}}{2(2)}$

$x=\frac{7\pm\sqrt{41}}{4}$

Therefore, The roots are $x=\frac{7+\sqrt{41}}{4},\frac{7-\sqrt{41}}{4}$