Question

Twice a number is added to three times it’s square. If the result is 16, find the number.

Answer 1

2 and - 2 2/3

Explanation:

`2x + {3x}^(2) = 16 \\ {3x}^(2) + 2x - 16 = 0 \\ D= {2}^(2) - 4 * 3 * ( - 16) = 4 * 192 = 196 \\ √(D) = √(196) = 14 \\ x = ( - 2 + 14)/(3 * 2) = (12)/(6) = 2 \\ x = ( - 2 - 14)/(3 * 2) = ( - 16)/(6) = - 2 * (2)/(3)`

Answer 2

Twice a number is added to three times it’s square. The number is 2, -2.67

SOLUTION:

Let the number be x.

Given, twice a number is added to three times it’s square.

twice a number + three times it’s square

2
`*` number + 3
`*` numbers square

`2 * x+3 *(x)^(2)`

`2 x+3 x^(2)`

Also given that, result is 16. So the above equation is equal to 16

`\begin{array}{l}{2 x+3 x^(2)=16} \\ {3 x^(2)+2 x-16=0}\end{array}`

Let us find roots of above equation using quadratic formula.

`x=\frac{-b \pm \sqrt{b^(2)-4 a c}}{2 a}`

Here, a = 3, b = 2, c = -16

`\mathrm{x}=\frac{-2 \pm \sqrt{2^(2)-4 * 3 *(-16)}}{2 * 3}`

On simplification we get,

`\begin{aligned} &=(-2 \pm √(4+12 * 16))/(6) \\ &=(-2 \pm √(196))/(6) \\ &=(-2 \pm 14)/(6) \\ &=(-2+14)/(6), (-2-14)/(6) \\ &=(12)/(6), (-16)/(6) \end{aligned}`

x = 2, -2.67

Hence, the value of "x" is either 2 or -2.67