The sign of the discriminant value tells us the number of real solutions the quadratic equation has. However, it cannot tell us by itself what the solutions are. We need to use the Quadratic Formula for this.
x=2a-b±b2−4ac
Notice that the discriminant is part of the formula.
Constructing a Counterexample
To give an counterexample to the exercise's statement, notice that we can have two different equations by swapping the values of a and c. Since the discriminant depends on their product it would still be the same for both, but the solutions will not be the same. We can start by writing two equations in the standard form a2+bx+c=0.
Notice that Equation II will have the same discriminant value, as b is the same for both equations, and the product 4ac would be same as well.
Finding the Solutions Using the Quadratic Formula
Now we can use the Quadratic Formula for this case, to show that even though both equations have the same discriminant values as we said before, they will have different solutions. Recall that we know that the discriminant is 4 in both cases.
Let's find the solutions for both equations by substituting the corresponding values into the above formula.
Equation (I)
Equation (II)
x=2(2)-6±4
x=2(4)-6±4
x=4-6±2
x=8-6±2
x1=4-6+2
x2=4-6−2
x1=8-6+2
x2=8-6−2
x1=-1
x2=-2
x1=-21
x2=-1
We can see that the solutions are not the same. For Equation (I) the solutions are x1=-1 and x2=-2, while Equation (II) has x1=-21 and x2=-1 for solutions. Notice that there are infinitely many solutions satisfying the exercise's requirements. This is only an example solution.
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