How do you solve #x^2 + 5x + 7 = 0# using the quadratic formula?
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#(-5+isqrt(3))/2# and #(-5-isqrt(3))/2#
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.
#x=(-5+isqrt3)/2# or #(-5-isqrt3)/2#
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#
#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
