How do you solve #x^2 - 3x - 4 = 0# using the quadratic formula?

Question

Eli Assouline · Accepted Answer

Identify the coefficients #a#, #b# and #c# and substitute those values into the quadratic formula to find #x=-1# or #x=4#.

#x^2-3x-4 = 0# is of the form #ax^2+bx+c = 0#, with #a=1#, #b=-3# and #c = -4#. The quadratic formula gives us solutions: #x = (-b+-sqrt(b^2-4ac))/(2a)# #=(3+-sqrt(3^2-(4xx1xx-4)))/(2*1)# #=(3+-sqrt(9+16))/2# #=(3+-sqrt(25))/2# #=(3+-sqrt(5^2))/2# #=(3+-5)/2# That is #x=-1# or #x=4#

Abigail Allen · Answer

undefined To solve the quadratic equation $ x^2 - 3x - 4 = 0 $ using the quadratic formula, first identify the coefficients $ a = 1 $, $ b = -3 $, and $ c = -4 $. Then, apply the quadratic formula: $ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $. Plug in the values of $ a $, $ b $, and $ c $, and solve for $ x $.

$$ x = \frac{{-(-3) \pm \sqrt{{(-3)^2 - 4 \cdot 1 \cdot (-4)}}}}{{2 \cdot 1}} $$

$$ x = \frac{{3 \pm \sqrt{{9 + 16}}}}{{2}} $$

$$ x = \frac{{3 \pm \sqrt{{25}}}}{{2}} $$

$$ x = \frac{{3 \pm 5}}{{2}} $$

Therefore, the solutions for $ x $ are $ x = \frac{{3 + 5}}{{2}} $ and $ x = \frac{{3 - 5}}{{2}} $.

Simplify these expressions to find the solutions:

$$ x_1 = \frac{{3 + 5}}{{2}} = 4 $$

$$ x_2 = \frac{{3 - 5}}{{2}} = -1 $$

Hence, the solutions to the equation are $ x = 4 $ and $ x = -1 $.