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:
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
[tex]5 x+2(x)^{2}[/tex]
Also given that, result is 168. So the above equation is equal to 168
[tex]\begin{array}{l}{5 x+2 x^{2}=168} \\ {2 x^{2}+5 x-168=0}\end{array}[/tex]
Let us find roots of above equation using quadratic formula.
[tex]x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}[/tex]
Here, a = 2, b = 5, c = -168
Substitute the values in formula we get
[tex]x=\frac{-5 \pm \sqrt{5^{2}-4 \times 2 \times(-168)}}{2 \times 2}[/tex]
[tex]=\frac{-5 \pm \sqrt{25+8 \times 168}}{4}[/tex]
On simplification we get,
[tex]=\frac{-5 \pm \sqrt{1369}}{4}[/tex]
[tex]\begin{array}{l}{=\frac{-5 \pm 37}{4}} \\\\ {=\frac{-5+37}{4}, \frac{-5-37}{4}} \\\\ {=\frac{32}{4}, \frac{-42}{4}}\end{array}[/tex]
x = 8, -10.5
Hence, the value of x is either 8 or -10.5
