Question

The sum of a square of a number and one less than the number, equals 9 less than, the square of one more than the same number.

Find the number​

Answer 1

Answer:

let the number be x

=>xsqaure- x sqaure+x-2x-1-1+9=0

Answer 2

Question

The sum of a square of a number and one less than the number, equals 9 less than, the square of one more than the same number.  Find the number​.

$\rule{300}{1}$

Answer

Let that number be 'x'

Now we will need to solve this equation to find the number ⇒ $x^{2}+x-1=(x+1)^{2}-9$

Let's solve your equation step-by-step.

⇒ $x^{2}+x-1=(x+1)^{2}-9$

Step 1: Simplify both sides of the equation.

⇒ $x^{2}+x-1=x^{2}+2x-8$

Step 2: Subtract x² from both sides.

⇒ $x^{2}+x-1-x^{2}=x^{2}+2x-8-x^{2}$

⇒ $x-1=2x-8$

Step 3: Subtract 2x from both sides.

⇒ $x-1-2x=2x-8-2x$

⇒ $-x-1=-8$

Step 4: Add 1 to both sides.

⇒ $-x-1+1=-8+1$

⇒ $-x=-7$

Step 5: Divide both sides by -1.

⇒ $\frac{-x}{-1} = \frac{-7}{-1}$

⇒ $x=7$

Now let's verify if the number we require is 7.

⇒ $x^{2}+x-1=(x+1)^{2}-9$

⇒ $7^{2}+7-1=(7+1)^{2}-9$

⇒ $49+7-1=(8)^{2}-9$

⇒ $56-1=64-9$

⇒ $55=55$

∴ LHS = RHS

∴ The required number is 7.

$\rule{300}{1}$