How do you find the roots, real and imaginary, of #y= x^2 + 4x - 3 # using the quadratic formula?

Question

How do you find the roots, real and imaginary, of #y= x^2 + 4x - 3   # using the quadratic formula?

Isaiah Anthony · Accepted Answer

See explanation...

#x^2+4x-3# is of the form #ax^2+bx+c# with #a=1#, #b=4# and #c=-3#. This has zeros given by the quadratic formula: #x = (-b+-sqrt(b^2-4ac))/(2a)# #=(-4+-sqrt((-4)^2-(4*1*(-3))))/(2*1)# #=(-4+-sqrt(16+12))/2# #=(-4+-sqrt(28))/2# #=(-4+-sqrt(2^2*7))/2# #=(-4+-2sqrt(7))/2# #=-2+-sqrt(7)# Alternative Method The difference of squares identity can be written: #a^2-b^2=(a-b)(a+b)# Complete the square and use this with #a=x+2# and #b=sqrt(7)# as follows: #x^2+4x-3# #=x^2+4x+4-7# #=(x+2)^2-(sqrt(7))^2# #=((x+2)-sqrt(7))((x+2)+sqrt(7))# #=(x+2-sqrt(7))(x+2+sqrt(7))# Hence zeros: #x=-2+-sqrt(7)#

Eva Adamson · Answer

undefined To find the roots of $y = x^2 + 4x - 3$ using the quadratic formula:

1. Identify the coefficients $a$, $b$, and $c$ in the quadratic equation $y = ax^2 + bx + c$. In this case, $a = 1$, $b = 4$, and $c = -3$.

2. Substitute these values into the quadratic formula: $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$.

3. Plug in the values: 
   $x = \frac{{-4 \pm \sqrt{{4^2 - 4 \cdot 1 \cdot (-3)}}}}{{2 \cdot 1}}$.

4. Simplify inside the square root: 
   $x = \frac{{-4 \pm \sqrt{{16 + 12}}}}{{2}}$, 
   $x = \frac{{-4 \pm \sqrt{{28}}}}{{2}}$.

5. Further simplify the square root if possible: 
   $x = \frac{{-4 \pm \sqrt{{4 \cdot 7}}}}{{2}}$, 
   $x = \frac{{-4 \pm 2\sqrt{{7}}}}{{2}}$.

6. Divide each term by 2: 
   $x = \frac{{-4}}{{2}} \pm \frac{{2\sqrt{{7}}}}{{2}}$, 
   $x = -2 \pm \sqrt{{7}}$.

Thus, the roots of the equation $y = x^2 + 4x - 3$ are $x = -2 + \sqrt{7}$ and $x = -2 - \sqrt{7}$.