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:

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

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

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

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

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

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

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