Question

Five times a number is added to two times it’s square. If the result is 168, find the number.

Answer 1

let number be x.

by the question

5x+2(x^2)=168

2x^2+5x-168=0

2x^2 +(21-16)x-168=0

2x^2+21x-16x-168=0

2x^2-16x+21x-168=0

2x(x-8)+21(x-8)=0

(x-8)(2x+21)=0

either

x-8=0

so. x=8

or

2x+21=0

2x=-21

x=-21/2

therefore x=8 or x=-21/2

Answer 2

Five times a number is added to two times it’s square. The value of “x” is either 8 or -10.5

SOLUTION:

Let the number be “x”

Given, Five times a number is added to two times it’s square.

Five times a number + two times it’s square .Hence we get

5x + 2x square

`5 x+2(x)^(2)`

Also given that, result is 168. So the above equation is equal to 168

`\begin{array}{l}{5 x+2 x^(2)=168} \\ {2 x^(2)+5 x-168=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 = 2, b = 5, c = -168

Substitute the values in formula we get

`x=\frac{-5 \pm \sqrt{5^(2)-4 * 2 *(-168)}}{2 * 2}`

`=(-5 \pm √(25+8 * 168))/(4)`

On simplification we get,

`=(-5 \pm √(1369))/(4)`

`\begin{array}{l}{=(-5 \pm 37)/(4)} \\\\ {=(-5+37)/(4), (-5-37)/(4)} \\\\ {=(32)/(4), (-42)/(4)}\end{array}`

x = 8, -10.5

Hence, the value of x is either 8 or -10.5