Question

4 times a number is 21 less than the square of that number. Find the positive solution.

Answer 1

Let x = your number.

4x = x^2 - 21. Subtract 4x from both sides.

0 = x^2 - 4x - 21. Factor.

0 = (x - 7) (x + 3). Set both terms equal to zero.

(x + 3) = 0. Subtract 3 from both sides.

x = -3. This is the negative solution.

(x - 7) = 0. Add 7 to both sides.

x = 7. This is the positive solution

Answer 2

The positive solution of the given quadratic equation is x = 7

How to solve the quadratic equation?

We can define the variable x as "the number".

Then "4 times a number is 21 less than the square of that number."

can be written as:

4x = x^2 - 21

This is just a quadratic equation, we can rewrite this as:

x^2 - 4x - 21 = 0

The solutions of the quadratic equation are given by Bhaskara's formula:

$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4*1*(-21)} }{2*1} \\\\x = \frac{4 \pm 10 }{2}$

We only want the positive solution, which is:

x = (4 + 10)/2 = 7

If you want to learn more about quadratic equations, you can read:

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