Question

7 times a number is 8 less than the square of that number. Find the positive solution

Answer 1

Answer:

The number is 8

Step-by-step explanation:

Let

x ----> a number

we know that

The algebraic expression is equal to

$7x=x^2-8$

$x^2-7x-8=0$

solve the quadratic equation

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$x^2-7x-8=0$

so

$a=1\\b=-7\\c=-8$

substitute in the formula

$x=\frac{-(-7)\pm\sqrt{-7^{2}-4(1)(-8)}} {2(1)}$

$x=\frac{7\pm\sqrt{81}} {2}$

$x=\frac{7\pm9} {2}$

$x=\frac{7+9} {2}=8$

$x=\frac{7-9} {2}=-1$

therefore

The number is 8

Answer 2

Answer:

8

Step-by-step explanation: I got it correct