How do you solve #x^2 + 5x – 6 = 0#?

Question

Nathan Axel · Accepted Answer

The solutions are

#color(green)(x=-6#

# color(green)(x=1# # x^2+5x–6=0# We can Split the Middle Term of this expression to factorise it and thereby find solutions. In this technique, if we have to factorise an expression like #ax^2 + bx+ c#, we need to think of 2 numbers such that: #N_1*N_2 = a*c = 1*-6 = -6# and #N_1 +N_2 = b = 5# After trying out a few numbers we get : #N_1 = -1# and #N_2 =6# #6*-1 = -6#, and #6+(-1)= 5# # x^2+5x–6= x^2+6x-1x–6# #x(x+6)-1(x+6)=0# #(x+6)# is a common factor to each of the terms #color(green)((x+6)(x-1)=0# We now equate factors to zero: #x+6=0, color(green)(x=-6# #x-1=0, color(green)(x=1#

Savannah Anderson · Answer

undefined To solve the quadratic equation $ x^2 + 5x - 6 = 0 $, you can use the quadratic formula:

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

Where $ a = 1 $, $ b = 5 $, and $ c = -6 $.

Substituting these values into the formula:

$$ x = \frac{{-5 \pm \sqrt{{5^2 - 4 \cdot 1 \cdot (-6)}}}}{{2 \cdot 1}} $$

$$ x = \frac{{-5 \pm \sqrt{{25 + 24}}}}{{2}} $$

$$ x = \frac{{-5 \pm \sqrt{{49}}}}{{2}} $$

$$ x = \frac{{-5 \pm 7}}{{2}} $$

Therefore, the solutions are $ x = \frac{{-5 + 7}}{{2}} = 1 $ and $ x = \frac{{-5 - 7}}{{2}} = -6 $.