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Chapter 2: Problem 89

Suppose \(b\) and \(c\) are numbers such that the equation $$ x^{2}+b x+c=0 $$ has no real solutions. Explain why the equation $$ x^{2}+b x-c=0 $$ has two real solutions.

Short Answer

Expert verified
For the first equation, \(x^2 + bx + c = 0\), its discriminant is \(D_1 = b^2 - 4c\), and it is given that \(D_1 < 0\). Now, for the second equation, \(x^2 + bx - c = 0\), its discriminant is \(D_2 = b^2 + 4c\). To show that the second equation has two real solutions, we need to prove that \(D_2 > 0\). Since \(b^2 - 4c < 0\), adding \(8c\) to both sides gives \(b^2 + 4c > 0\), which means \(D_2 > 0\). Therefore, the second equation has two real solutions.

Step by step solution

01

Find the discriminant for the first equation

The discriminant of a quadratic equation \(ax^2 + bx + c = 0\) is given by \(D = b^2 - 4ac\). For the given first equation, \(a = 1\), \(b = b\), and \(c = c\). So we have: \[D_1 = b^2 - 4c\] It is given that the first equation has no real solutions, so the discriminant must be negative, that is: \[D_1 < 0\] \[b^2 - 4c < 0\]
02

Find the discriminant for the second equation

Now, let's find the discriminant for the second equation. In this case, everything remains the same, except for the constant term, which is now \(-c\). Therefore, we have: \[D_2 = b^2 - 4(-c)\]
03

Prove second discriminant is positive

Now we need to show that \(D_2\) is positive, which implies that the second equation has two real solutions. From our calculation above, we have: \[D_2 = b^2 + 4c\] Recall that from the first equation's discriminant, we have: \[b^2 - 4c < 0\] So, if we add \(8c\) to both sides, we get: \[b^2 + 4c > 0\] which is equivalent to: \[D_2 > 0\]
04

Conclusion

Since the discriminant \(D_2\) of the second equation is positive, the equation \(x^2 + bx - c = 0\) has two real solutions.

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