Problem 49 - Precalculus Student Solutions Manual 5th

Question

Francisco claimed that the equation \(x^{2}-8 x=0\) cannot be solved by the quadratic formula since there is no value for \(c .\) Is he correct?

Step-by-Step Solution

Answer

Francisco is incorrect; the equation can be solved using the quadratic formula.

Step 1 - Understand the Quadratic Formula

Recall the quadratic formula, which is used to solve quadratic equations of the form \(ax^2 + bx + c = 0\):\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Step 2 - Identify the Coefficients

Identify the coefficients \(a\), \(b\), and \(c\) in the given equation \(x^2 - 8x = 0\). By comparing it with the standard form \(ax^2 + bx + c = 0\), we get:- \(a = 1\)- \(b = -8\)- \(c = 0\)

Step 3 - Substitute into the Quadratic Formula

Substitute the identified coefficients into the quadratic formula:\[ x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot 0}}{2 \cdot 1} \]This simplifies to:\[ x = \frac{8 \pm \sqrt{64}}{2} \]

Step 4 - Solve for x

Solve the equation:\[ x = \frac{8 \pm 8}{2} \]This gives two solutions:\[ x = \frac{8 + 8}{2} = 8 \]\[ x = \frac{8 - 8}{2} = 0 \]Therefore, the solutions are \(x = 8\) and \(x = 0\).

Step 5 - Conclusion

Even though \(c = 0\), the quadratic formula can still be used. Francisco's claim that the equation cannot be solved using the quadratic formula is incorrect.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool in algebra for solving equations of the form \(ax^2 + bx + c = 0\). The formula itself is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] It allows us to find the values of x that satisfy the equation by substituting the coefficients \(a\), \(b\), and \(c\). This formula works for all quadratic equations, whether they have real or complex solutions.
Using this formula correctly requires identifying the right coefficients and substituting them properly.

Coefficients Identification

Identifying coefficients in a quadratic equation is a crucial first step to using the quadratic formula. The general form of a quadratic equation is \(ax^2 + bx + c = 0\). Here, \(a\) is the coefficient of \(x^2\), \(b\) is the coefficient of \(x\), and \(c\) is the constant term.

• For the given equation \(x^2 - 8x = 0\)
• We see that \(a = 1\)
• \(b = -8\)
• There is no constant term, so \(c = 0\).

Identifying these correctly ensures that we substitute the right values into the quadratic formula.

Algebraic Solution Steps

Once we have identified the coefficients, we substitute them into the quadratic formula. For the equation \(x^2 - 8x = 0\):

• We have \(a = 1\), \(b = -8\), and \(c = 0\).

Substituting these into the formula:
\(x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot 0}}{2 \cdot 1} \):
This simplifies to:
\(x = \frac{8 \pm \sqrt{64}}{2} \):
Further simplifying, we solve for \(x\):
\(x = \frac{8 + 8}{2}=8\)
\(x = \frac{8 - 8}{2}= 0\).
So, our solutions are \(x = 8\) and \(x = 0\).
Following these algebraic steps carefully allows us to find the correct values of \(x\).

Zero Coefficient

Sometimes one of the coefficients in a quadratic equation could be zero. In our original equation \(x^2 - 8x = 0\), the coefficient \(c\) is zero.

• This simplifies our calculations because the term involving \(c\) in the quadratic formula becomes zero.

This does not mean the quadratic formula is unusable. It still provides valid solutions.
When substituting into \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), the term \(-4ac=0\), simplifying our square root term significantly. Even with a zero coefficient, the quadratic formula efficiently finds the solutions.