How do you solve 4x^2-x+5=0?

2 Answers
Nov 18, 2016

The answer is =(1+-isqrt79)/8

Explanation:

The equation is ax^2+bx+c=0

4x^2-x+5=0

Let's calculate the discriminant,

Delta=b^2-4ac=1-4*4*5=-79

Delta<0

So the solutions are not in RR but in CC

x=(-b+-sqrtDelta)/(2a)

x=(1+-isqrt79)/8

where i^2=-1

graph{4x^2-x+5 [-24.46, 21.15, -2.57, 20.25]}

Nov 18, 2016

x=1/8 \pm (i sqrt(79))/8

Explanation:

If you have an equation in the form of ax^2+bx+c=0 you can apply the quadratic formula

x=(-b\pm sqrt(b^2 -4ac))/(2a)

Where a is the coefficient of x^2 and b is the coefficient of x and c is the constant

4x^2-x+5=0

In this example a=4 and b=-1 and c=5

x=(-(-1)\pm sqrt((-1)^2-4(4)(5)))/(2(4))

=(1\pm sqrt(-79))/8=1/8 \pm sqrt(-79)/8

You can not have a negative under the square root, we can use the definition that i=sqrt(-1) to extract the negative

1/8 \pm sqrt(-79)/8=1/8 \pm (i sqrt(79))/8

The final answer is x=1/8 \pm (i sqrt(79))/8