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

1 Answer
Apr 11, 2018

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