Which method do you use to solve #7x^2 + 8x + 100#?

Question

Kennedy Adams · Accepted Answer

A quadratic equation is simply another way of solving a problem if the solution cannot be factored logically.  First we can start with some quick review: Let’s say we have the equation #x^(2)+ 2x - 3# for example. This equation could be solved logically using the factors of the first and last terms.  To begin, we can state the factors of the first term, #x^(2)#. Imagine there’s an invisible 1 in front of the #x^(2)#, therefore the factors are 1, because only #1 * 1, or -1*-1# will multiply to get one. Then we can analyze the third term, #-3#. The factors of #-3# are either #1 * -3, or -1 * 3#.  Now we can check and see if any of the factors can combine in order to get a #+2#, the middle term (don’t worry about the x’s, those will carry over). Recall #1= -1, 1#, and #-3 = 1, -1, 3, -3# From our factors we can use a -1 and a 3 to get +2.  Therefore, 
#(x+3)(x-1)=0# is our derived factorization. Then plug in the values to make the statement true, -3 or 1 will both result in an answer of 0 and our the possible values for x.  However , when the logical factorization seen above is not possible, we can plug our numbers into the quadratic equation . #ax^(2)+bx+c# is the standard way we view an equation. Using the values from the equation above, #a= 1, b=2, and c=-3#. After our a, b, and c values are found we can plug them into the actual quadratic equation.  #(-b+-sqrt(b^(2)-4ac))/(2a)# Note : This equation may look intimidating, but as long as you follow factoring rules, you should have no problem. It’s totally normal to come out with an answer containing square roots.

Genesis Allen · Answer

We do not "solve" expressions.
Perhaps you want to know how to solve the equation #7x^2+8x+100=0#.  .
Or perhaps you want to know how to factor the expression #7x^2+8x+100# In order to do the second for this expression, we'll have to do the first.  That's because this expression is irreducible using real, rational coefficients. Solve:
#7x^2+8x+100=0#.  . Spend a little time factoring, but not too much, because we know we can always solve by the quadratic formula. #x=(-b+-sqrt(b^2-4ac))/(2a)# For the equation we're working on:
#x=(-(8)+-sqrt((8)^2-4(7)(100)))/(2(7))#  so  #x=(-8+-sqrt(64-2800))/14=(-8+-sqrt(-2736))/14#
#=(-8+-12sqrt19i)/14=(-4)/7 +- (6sqrt(19))/7i# Note
Simplifying sqrt(2736) is not trivial. #2736# is divisible by #9# (the digits add up to 18, which is divisible by 9) #2736=9*304#
#304# is divisible by #4# (half or 304 is still even, so we can divide again) #2736=9*4*76#  (half of 76 is 38, still even)
#2736=9*4*4*19#  So #sqrt(2736)=sqrt(9*4*4*19)=3*2*2*sqrt19=12sqrt19# Now that we know the zeros of #7x^2+8x+100# are #z_1# and #z_2#, we can factor the quadratic as #(x-z_1)(x-z_2)#, so: #7x^2+8x+100=(x-((-4)/7 + (6sqrt(19))/7i))(x-((-4)/7 - (6sqrt(19))/7i))#

Genesis Allen · Answer

undefined To solve the quadratic equation $7x^2 + 8x + 100$, you can use the quadratic formula:

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

where $a = 7$, $b = 8$, and $c = 100$. Plugging these values into the formula, you can calculate the solutions for $x$.