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 $1 \frac{1}{2}$ $1 1/2$ or $- 2$ $-2$ .

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

$2 x^{2} + x = 6$ $2x^2+x=6$

Subtract $6$ $6$ from both sides.

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

Factorise.

$2 x^{2} + 4 x - 3 x - 6 = 0$ $2x^2+4x-3x-6=0$

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

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

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

$x = \frac{3}{2} = 1 \frac{1}{2}$ $x=3/2=1 1/2$ or $x = - 2$ $x=-2$

Answer 2 · 2017-01-11T04:47:20

I get two numbers: $x = - 2, \frac{6}{4} = \frac{3}{2}$ $x=-2, 6/4=3/2$

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

If a number:

$x$ $x$

is added to twice it's square:

$x + 2 x^{2}$ $x+2x^2$

the result is 6:

$x + 2 x^{2} = 6$ $x+2x^2=6$

Now let's find the number:

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

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

$x = \frac{- 1 \pm \sqrt{1^{2} - 4 (2) (- 6)}}{2 (2)}$ $x=(-1+-sqrt(1^2-4(2)(-6)))/(2(2))$

$x = \frac{- 1 \pm \sqrt{1^{2} + 48}}{4}$ $x=(-1+-sqrt(1^2+48))/(4)$

$x = \frac{- 1 \pm \sqrt{49}}{4}$ $x=(-1+-sqrt(49))/(4)$

$x = \frac{- 1 \pm 7}{4}$ $x=(-1+-7)/(4)$

$x = - 2, \frac{6}{4} = \frac{3}{2}$ $x=-2, 6/4=3/2$