If a number is added to twice its square, the result is 6. How do you find the number?

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Answer 1 · 2017-01-11T04:32:52

The number could be #1 1/2# or #-2#.

From the data, taking the number to be #x#, we write:

#2x^2+x=6#

Subtract #6# from both sides.

#2x^2+x-6=0#

Factorise.

#2x^2+4x-3x-6=0#

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

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

#2x-3=0# or #x+2=0#

#x=3/2=1 1/2# or #x=-2#

Answer 2 · 2017-01-11T04:47:20

I get two numbers: #x=-2, 6/4=3/2#

Let's have the unknown number be #x#.

If a number:

#x#

is added to twice it's square:

#x+2x^2#

the result is 6:

#x+2x^2=6#

Now let's find the number:

#2x^2+x-6=0#

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#x=(-1+-sqrt(1^2-4(2)(-6)))/(2(2))#

#x=(-1+-sqrt(1^2+48))/(4)#

#x=(-1+-sqrt(49))/(4)#

#x=(-1+-7)/(4)#

#x=-2, 6/4=3/2#