Question

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

Answer 1

Answer:

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

ANSWER:

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

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

$=\frac{-5 \pm \sqrt{25+8 \times 168}}{4}$

On simplification we get,

$=\frac{-5 \pm \sqrt{1369}}{4}$

$\begin{array}{l}{=\frac{-5 \pm 37}{4}} \\\\ {=\frac{-5+37}{4}, \frac{-5-37}{4}} \\\\ {=\frac{32}{4}, \frac{-42}{4}}\end{array}$

x = 8, -10.5

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