Liesl is correct, if the discriminant is a perfect square, then we can factorize the quadratic equation.
Is the statement true or false?
For a quadratic equation:
y = a*x^2 + b*x + c
The discriminant is:
D = b^2 - 4ab
Notice that for the equation:
0 = a*x^2 + b*x + c
The solutions are given by:
[tex]x = \frac{-b \pm \sqrt{D}}{2a}[/tex]
If D is a perfect square, then we know that the square root has solutions, so there are two values:
[tex]x_1 = \frac{-b + \sqrt{D}}{2a}\\\\x_2 = \frac{-b - \sqrt{D}}{2a}[/tex]
Now with these solutions, we can factorize our equation as:
[tex]y = a*(x - ( \frac{-b + \sqrt{D}}{2a}))*(x - ( \frac{-b - \sqrt{D}}{2a}))[/tex]
So Liesl is correct, if the discriminant is a perfect square, we can factorize the quadratic equation.
If you want to learn more about factorization, you can read:
https://brainly.com/question/11579257
