How do you solve #5x^2-34x-7=0#?

1 Answer
Apr 11, 2018

Answer:

#x=7# and #x=-1/5#

Explanation:

#1)# Multiply the coefficient of x by the constant.

#5xx7=35#

#2)# Find a pair of numbers that add to make the second term and multiply to make the third. We can easily spot here that only #35# and #1# can make #34#

#3)# Work out which is negative and which is positive if necessary, as the second term is #-34x#, the numbers must be #-35x# and #+x#

Keep the quadratic constant at it's original, we just need to multiply to find the pair of numbers

#5x^2-34x-7 -> 5x^2+x-35x-7#

#4)# Factor out the first #2# terms:

#5x^2+x -> x(5x+1)#

#5)# Factor out the second #2# terms.

#-35x-7 -> -7(5x+1)#

Notice that both brackets are the same

So the first bracket is #(5x+1)#

#6)# The second bracket is the other #2# terms.

We have:

#x(5x+1)-7(5x+1)#

#therefore# The other bracket must be #(x-7)#

7) Solve:

#(5x+1)(x-7)=0#

#8)# Solve each bracket separately:

#5x+1=0#

#-> 5x=-1#

#-> x=-1/5#

#x-7=0#

#-> x=7#

Therefore these are our two answers.

#x=7# and #x=-1/5#

Remember you can always expand the brackets to check the answer