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

Question

Hazel Anglin · Accepted Answer

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

For the following type of quadratic equations:

#ax^2+bx+c#

Here is the quadratic formula:

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

Using the provided equation, we obtain:

#bba =1#

#bb(b)=5#

#bbc=7#

Utilizing the quadratic formula with these values:

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

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

This can be expressed as follows:

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

If #sqrt(-1)=i#

Then:

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

We refer to these as complex roots.

Ethan Adkins · Answer

#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#

Eli Assouline · Answer

No real solutions

#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# Not a true fix

Genesis Allen · Answer

undefined To solve the quadratic equation $x^2 + 5x + 7 = 0$ using the quadratic formula, first identify the coefficients:

$a = 1$, $b = 5$, and $c = 7$.

Then, substitute these values into the quadratic formula:

$x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$.

Substitute the values of $a$, $b$, and $c$:

$x = \frac{{-5 \pm \sqrt{{5^2 - 4(1)(7)}}}}{{2(1)}}$.

Calculate inside the square root:

$x = \frac{{-5 \pm \sqrt{{25 - 28}}}}{{2}}$.

$x = \frac{{-5 \pm \sqrt{{-3}}}}{{2}}$.

Since the discriminant $b^2 - 4ac$ is negative, the roots will be complex numbers.

$x = \frac{{-5 \pm i\sqrt{{3}}}}{{2}}$.