solve this equation (3x-7)(x+5)=0?

Question

what numbers can x be

Answer 1

x=7/3 or -5

Answer 2

Well, look at it. What number can I put into one of those to make it zero?

If you found a number to make 3x - 7 become zero, or x + 5 become zero, the left hand side of your equation will be zero.

The first one is made zero by x=7/3, and the other by x=-5. These two are possible solutions.

Hope that helped,
UMRmathmajor

Answer 3

7/3 or -5

Answer 4

x= -5

Answer 5

[3x-7][x+5]=0
3x2+15x-7x-35=0
3x2+8x-35=0
ok so this is expanding it. you just have to go vice versa when you get the equ 3x2+8x-35=0. what you got is the simplified form already and that is all you need to get x. just think about it - what do you have to multiply with to get a 0?..well if you dont know that would be a 0...so you just have to make one of the brackets equal 0:
3x-7=0
3x=7
x=7/3

AND

x+5=0
x=-5

because this is a quadratic equation, it has 2 valid values for x and therefore for this equ, they are:
x=7/3 AND x=-5

Answer 6

that is easy, anything times 0 is 0 so find a number that you would add to 5??..

so x is -5

-5+5=0 (3*-5=-15)-7 = -22
so (-22)(0)=0
simple

Answer 7

you need to use the rational root theorem to get conceivable rational roots of +/-{a million,5}. bear in techniques, to get those numbers... aspects of the consistent term (5) divided by potential of aspects of ideal coefficient (a million). attempt values to be sure -a million is a root (answer), so x - (-a million) = (x+a million) is a factor. divide to get x^3 + 3x^2 + 7x + 5 = (x+a million)(x^2 + 2x + 5) so (x+a million)(x^2 + 2x + 5) = 0. use quadratic formulation (or polishing off the sq. to remedy the three term factor. you at present have the three recommendations (and the form): x = -a million-2i, -a million, -a million+2i (the place the ordinary distinction is 2i.

Answer 8

1st step -> (3x-7)=0 and (x+5)=0
2nd step -> 2x=7 and x= -5
3rd step is x=7/2
therefore, answer of x can be 7/2 or -5

Answer 9

-5

Answer 10

3x7x+5=63