How do you solve #(r-2)(r+6)=4# using the quadratic formula?

Question

Eva Adamson · Accepted Answer

Solution: #r=2.47(2dp) or r= -6.47(2dp)# #(r-2)(r+6)=4 or r^2+4r-12=4 or r^2+4r-16=0 or (r+2)^2-4-16=0 or (r+2)^2=20 or r+2 = +-sqrt(20) or r+2=+-2sqrt5 :. r= -2+2sqrt5=2(-1+sqrt5)=2.47(2dp) =or r= -2-2sqrt5= 2(-1-sqrt5)=-6.47(2dp)# Solution: #r=2.47(2dp) or r= -6.47(2dp)# Using Quadratic formula:#(r=-b/(2a)+-sqrt(b^2-4ac)/(2a))# #r^2+4r-16=0 ; a=1 ;b= 4 ;c=-16 ; b^2-4ac=80 ; r= -4/2+- sqrt80/2= -2+-2sqrt5:. r_1=2.47(2dp) or r_2=-6.47(2dp)# [Ans]

Eli Assouline · Answer

undefined To solve the quadratic equation (r - 2)(r + 6) = 4 using the quadratic formula, first, expand the left side of the equation to get r^2 + 4r - 12 = 4. Then, move all terms to one side to form the standard quadratic equation format, which is r^2 + 4r - 16 = 0. Next, identify the coefficients a, b, and c in the quadratic equation ax^2 + bx + c = 0. In this case, a = 1, b = 4, and c = -16. Substitute these values into the quadratic formula:

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

$$ r = \frac{{-4 \pm \sqrt{{4^2 - 4(1)(-16)}}}}{2(1)} $$

$$ r = \frac{{-4 \pm \sqrt{{16 + 64}}}}{2} $$

$$ r = \frac{{-4 \pm \sqrt{{80}}}}{2} $$

$$ r = \frac{{-4 \pm 4\sqrt{5}}}{2} $$

$$ r = -2 \pm 2\sqrt{5} $$

Therefore, the solutions to the quadratic equation (r - 2)(r + 6) = 4 are $ r = -2 + 2\sqrt{5} $ and $ r = -2 - 2\sqrt{5} $.