How does square root property and real numbers relate to quadratic equations?

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ishpiro | College Teacher | (Level 1) Educator

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To solve quadratic equations, it is necessary to use square roots. For example, to solve the equation

`x^2 = 4`

one has to take square root of both sides:

`sqrt(x^2) = sqrt(4)`

The property of square root is that `sqrt(x^2) = |x|`

|x| = 2

`x = +-2` 

(x can be positive or negative 2 to satisfy the equation `x^2 = 4` ) This quadratic equation has 2 real solutions.

For more complicated quadratic equations, such as `x^2 + 6x +8=0` , the solution procedure is longer but the principle remains the same. This quadratic equation also has two real solutions.

But the quadratic equation such as

`x^2 =-4`

does not have any real solutions because it is impossible to take a square root of a negative number:

`sqrt(-4) ` is not a real number.

For general quadratic equation in the form `ax^2 + bx + c = 0`

the solutions are given by quadratic formula:

`x_(1,2) = (-b+-sqrt(b^2 - 4ac))/(2a)`

The expression under the radical, `b^2 - 4ac` , is called "discriminant". If it is positive, the quadratic equation has two real solutions. If it 0, the equation has 1 real solution. If it is negative, the equation has no real solutions.

 

 

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