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

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Answer 1 · 2016-11-18T14:10:50

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

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]}

Answer 2 · 2016-11-18T14:32:05

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

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#