How do you solve #x^2 + 5x + 7 = 0# using the quadratic formula?

3 Answers
Mar 22, 2018

Answer:

#(-5+isqrt(3))/2# and #(-5-isqrt(3))/2#

Explanation:

For quadratic equations of the form:

#ax^2+bx+c#

The quadratic formula is given by:

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

From given equation we have:

#bba =1#

#bb(b)=5#

#bbc=7#

Putting these values in the quadratic formula:

#(-(5)+-sqrt((5)^2-4(1)(7)))/(2(1))=(-5+-sqrt(25-(28)))/(2)#

#=(-5+-sqrt(-3))/2#

We can write this in the following way:

#sqrt(-3)=sqrt(3xx-1)=sqrt(3)*sqrt(-1)#

If #sqrt(-1)=i#

Then:

#(-5+isqrt(3))/2# and #(-5-isqrt(3))/2#

These are known as complex roots.

Mar 22, 2018

Answer:

#x=(-5+isqrt3)/2# or #(-5-isqrt3)/2#

Explanation:

Accordng to quadratic formula, solution of quadratic equation #ax^2+bx+c=0# is

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

Hence solution of #x^2+5x+7=0# is

#x=(-5+-sqrt(5^2-4*1*7))/2#

= #(-5+-sqrt(25-28))/2#

= #(-5+-sqrt(-3))/2#

i.e. #x=(-5+isqrt3)/2# or #(-5-isqrt3)/2#

Mar 22, 2018

Answer:

No real solutions

Explanation:

#x^2+5x+7=0#

Use the quadratic formula with #a=1, b=5, c=7#

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

#x=(-5+-sqrt((5)^2-4(1)(7)))/((2)(1))#

#x=(-5+-sqrt(25 - 28))/2#

#x=(-5+-sqrt(-3))/2#

No real solution