can you please show the work for: 2x^2 - x - 3 = 0 ?

Question

Thank you

Answer 1

x = 1.5, 1.0

Answer 2

2x^2-x-3 = 0

all u need to do is put it in a general form = ax^2+bx+c = 0

as it is in that form at this moment, u factorise it



2x^2-x-3 = 0

(2x-3)(x+1) = 0

so, u find x



how to find it?

u know that in linear equation watever u do on one side, u do the same on all side right? well,u 1st divided by the 1st bracket then the second. so, u get



(2x-3)(x+1)/x+1=0/x+1

so, u get 2x-3 = 0

move all the numbers, u get

2x-3=0

2x=3

x=3/2 or 1.5



u left one more. now u do the same thing but for the other bracket. so u get

x+1 = 0

x = -1 (is rejected as normally negative value is not accepted)



but, for the answer u give,0.5, it is natural to assume that u take the roots of x and add them.so, u get,

x= 1.5 + (-1)

= 1.5 - 1

= 0.5



thanks

Answer 3

simply factor:

(2x-3)(x+1) = 0



2x=3 x=-1

x= 3/2, x=-1

Answer 4

this is a second degree quadratic in x of the form

ax^2+bx+c=0

roots are given by

(-b+sqrt(b^2-4ac) ) /2a or (-b-sqrt(b^2-4ac)) /2a

(1+sqrt(1+24))/2*2 or (1-sqrt(1+24))/2*2

x=1.5 x=-1

Answer 5

2x^2-x-3=0

2x^2-3x+2x-3 = 0

x(2x-3)+1(2x-3)=0

(2x-3)(x+1)=0





2x-3=0

2x=3

x=3/2

x=1.5





x+1=0

x=-1







So:

x=1.5

x=-1