Question

Solve by factoring and list only the positive solution: 2x^2 - 5x = 88

Answer 1

can u factor it ?

Answer 2

Doesn't it have to equal 0?

Answer 3

Would it be 2x2 - 5x - 88=0? first

Answer 4

That's correct.
The first step is to put everything on the left side equaling zero like you did.
Now you need to factor it.

Answer 5

or you can also do middle term breaking.

Answer 6

Can you explain more please?

Answer 7

How would I factor 2x^2 - 5x - 88=0??

Answer 8

Find the solution of the equal - find discriminant D=5^2-2*4*(-88)=.. can you find roots if you know discriminant? Roots are 8 and -11/2. so factoring will give you (x-8)*(x+11/2)=0

Answer 9

@Keridavis

You need to factor this polynomial equation:
\(2x^2 - 5x - 88=0\)

This is a polynomial of the type \(ax^2 + bx + c\) , where \(a \ne 1\).

This is how you factor this type of polynomial:
1. Multiply ac together.
In your case, a = 2, and c = -88, so ac = -176.

2. Find two numbers that multiply to ac and add to b.
In your case, ac = -176, and b = -5
Since 176 = 2 * 2 * 2 * 2 * 11, we can see that -16 * 11 = -176, and
-16 + 11 = -5, so our two numbers are -16 and 11.

3. Break up the middle term of the polynomial using the two numbers found in the step above.

\(2x^2 - 5x - 88=0\)
becomes:
\(2x^2 - 16x + 11x - 88 =0\)

4. Factor by grouping. That means factor out a common factor from the first two terms, and factor out a common factor from the last two terms.

\(2x^2 - 16x + 11x - 88 =0\)
after factoring by grouping we get:
\(2x(x - 8) + 11(x - 8)=0\)

Notice we have the term (x - 8) in common, so we can factor it out leaving:

\((x - 8)(2x + 11) = 0\)

Now that the trinomial is factored, you can continue to solve for x by letting each factor equal zero.

\(x - 8 = 0\) or \(2x + 11 = 0\)

\(x = 8\) or \(2x = -11\)

\(x = 8\) or \(x = -\dfrac{11}{2} \)

Since you are asked to only show positive solutions, your answer is x = 8.

Answer 10

Do you understand the method?