How do you solve 5x2-13x+6 Also how do you solve x2-36
First, set 2 sets of parentheses like so: ( )( )
In order to get `5x^2` we must break into factors. The only 2 factors that could work are `5x` and x.
Therefore we have: (5x - )(x - ). Both signs must be negative as middle term is negative and last is positive. Next we need 2 terms that multiply to equal 6. This can only be either 2 and 3 or 6 and 1.
Therefore we could have:
(5x-6)(x-1) or (5x-1)(x-6) or (5x-2)(x-3) or (5x-3)(x-2)
In order to determine the correct solution, you must find the sum the product of inner and outer terms.
Case 1 gives: -5x and -6x = -11x
Case 2 gives: -30x and -1x = -31x
Case 3 gives: -15x and -2x = -17x
Case 4 gives: -10x and -3x = -13x
The last instance is correct:
The solution: `(5x-3)(x-2)`
Solve `x^2-36` .
In this case you have no middle term, and the 2 given terms are perfect squares. This is a special case known as the difference of 2 squares.
( )( ) Same start, however keep in mind there is no middle term.
(x )(x ) to make the `x^2`
Next, since 36 is a perfect square identify the square root, which is 6.
(x 6)(x 6), since we want the middle term to "cancel out" each sign will be opposite.
`(x+6)(x-6)` Notice outer and inner terms are 6x and -6x which equal 0
(x-6)(x+6) is your solution.
How do you solve `5x^2-13x+6` ( `ax^2+bx+c` )
The first step of solving this is by multiplying a by c
`5 xx 6 = 30`
Then you find factors of 30 that add up to -13 (3 and 10) put those as b
`5x^2 - 10x - 3x + 6`
`(5x^2 - 10x) (- 3x + 6)`
`5x (x - 2) -3 (x - 2 )`
`(5x - 3) (x - 2)`
the solutions are x = 3/5 and x = 2
Also how do you solve x2-36
For this one we use the difference of 2 squares (a - b)(a + b)
(x - 6) (x + 6)
=x^2 - 6^2
we know that a^2 - b^2 =(a+b)(a-b), so
Note ^ symbol used for power of the exponent .
I hope, it will really helpful to you.