problem help?

Question

i feel retarted for asking this because i know i can do it..im just drawin a blank.

2x^2 + 14x = 88

Answer 1

OK, first off get everything over to the left hand side so that when we start dividing out we have zero on the other side.

2x^2 + 14x - 88 = 0

Divide through by two (this simplifies things since we don't have to worry about 0/2, 0/n will always be 0)

x^2 + 7x - 44 = 0

Then factorise out (we need two numbers which when added together = 7 and when multiplied = 44)

(x + 11) (x - 4) = 0

This means that x + 11 = 0 and x - 4 = 0.

So x = -11 or +4

Plug these back into the original equation to validate:

(2 * -11 * -11) + ( 14 * -11) = (2 * 121) - 154 = 242 - 154 = 88

or

( 2 * 4 * 4) + (14 * 4) = (2 * 16) + 56 = 32 + 56 = 88.

Answer 2

*Set the equation to "0" --- subtract 88 from both sides (when you move a term to the opposite side, always use the opposite sign)....

2x^2 + 14x - 88 = 88 - 88

2x^2 + 14x - 88 = 0

First: factor the expression --- find a number divisible with the coefficients, which is 2....

2(x^2 + 7x - 44) = 0

Sec: factor the expression in parenthesis > multiply the 1st & 3rd coefficient to get "-44." Find two numbers that give you "-44" when multiplied & "7" (2nd/middle coefficient) when added/subtracted. The numbers are (11 & - 4).

*Rewrite the expression with the new middle coefficients...

x^2 + 11x - 4x - 44

*With 4 terms - group "like" terms & factor both sets of parenthesis....

(x^2 - 4x) + (11x - 44)
x(x - 4) + 11(x - 4)
(x - 4)(x + 11)

Now, you have >>>

2(x - 4)(x + 11) = 0

Third: solve the "x" variables --- set both parenthesis to "0"....

a. x - 4 = 0

*Isolate "x" on one side --- add 4 with both sides....

x - 4 + 4 = 0 + 4
x = 0 + 4
x = 4

b. x + 11 = 0

*Isolate "x" on one side --- subtract 11 from both sides....

x + 11 - 11 = 0 - 11
x = 0 - 11
x = -11

Solutions: -11 and 4

Answer 3

2x² + 14x = 88

2x² + 14x - 88 = 88 - 88

2x² + 14x - 88 = 0

x² + 7x - 44 = 0 . . . . Diving through by 2

x² + 11x - 4x - 44

x(x + 11) -4(x + 11)

(x - 4)(x + 11)

- - - - - - - - - - - - -s-

Answer 4

2x² + 14x = 88
2(x² + 7x - 44) = 0
D = 7² - 4.1.-44
D = 49 + 176
D = 225

x = (-7 +/- \/225) : 2.1
x' = (-7 + 15) : 2 = 4
x" = (-7 - 15) : 2 = -11
(There are x and y in R| x' = -11 x" = 4}
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Answer 5

use the quadratic formula
where a=2, b=14, c=-88
you'll get: x = 4 and x = -11
done

Update: If you use the quadratic formula, you will get an answer even when you can't see how to factor your problem so easily. It always works.