How do you solve #x^2 + x +10 = 0# using the quadratic formula?

3 Answers
Apr 3, 2016

Answer:

Use the quadratic formula to find roots:

#x =-1/2+-sqrt(39)/2 i#

Explanation:

#x^2+x+10 = 0# is of the form #ax^2+bx+c = 0# with #a=1#, #b=1# and #c=10#.

It has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = 1^2-(4*1*10) = 1-40 = -39#

Since this is negative, this quadratic equation has no Real roots.

It has a Complex conjugate pair of roots given by the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#=(-b+-sqrt(Delta))/(2a)#

#=(-1+-sqrt(-39))/2#

#=(-1+-sqrt(39)i)/2#

#=-1/2+-sqrt(39)/2 i#

Apr 3, 2016

Answer:

zero

Explanation:

#x^2+x+10=0# is in the form of #ax^2+bx+c=0#. Then you apply the quadratic formula
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You type into the calculator this:

x=-1+√#1^2#- 4110/2*1

This equals to zero because a surd cannot be a negative number. (Some equations cannot be solved and this is one of them)

Apr 4, 2016

Answer:

#x=-1/2+-(sqrt39)/2i#

Explanation:

#color(blue)(x^2+x+10=0#

This is a Quadratic equation (in form #ax^2+bx+c=0#)

Use Quadratic formula

#color(brown)(x=(-b+-sqrt(b^2-4ac))/(2a)#

Remember that #a,band c# are the coefficients of #x^2,xand10#

Where,

#color(red)(a=1,b=1,c=10#

And don't be afraid with the formula!

#rarrx=(-1+-sqrt(1^2-4(1)(10)))/(2(1))#

#rarrx=(-1+-sqrt(1-4(10)))/(2)#

#rarrx=(-1+-sqrt(1-40))/(2)#

#rarrx=(-1+-sqrt(-39))/(2)#

Oh! we cannot find the square root of #-39# because it is a negative number!.Don't worry,in such cases it is called a complex number
(in form #sqrt(-1)#) .It is represented as #i#

So,

#rarrx=(-1+-sqrt(39*-1))/(2)#

#rarrx=(-1+-sqrt39)/(2)i#

#color(green)(rArrx=(-1)/(2)+-(sqrt39)/(2)i#