Question

If \(\alpha, \beta\) are the roots of \(a x^{2}+b x+c=0\) and \(\alpha+k, \beta+k\) are the roots of \(p x^{2}+q x+r=0\), then \(k=\) (a) \(-\frac{1}{2}\left(\frac{a}{b}-\frac{p}{q}\right)\) (b) \(\left(\frac{a}{b}-\frac{p}{q}\right)\) (c) \(\frac{1}{2}\left(\frac{b}{a}-\frac{q}{p}\right)\) (d) \((a b-p q)\)

Short answer

Answer: \(k=\frac{1}{2}\left(\frac{b}{a}-\frac{q}{p}\right)\)

Step-by-step solution

Write down the relationships for quadratic equation roots

For the first quadratic equation \(a x^{2} + b x + c = 0\), the roots are \(\alpha\) and \(\beta\). By Vieta's formulas, we have: Sum of the roots: \(\alpha+\beta=-\frac{b}{a}\) Product of the roots: \(\alpha\beta=\frac{c}{a}\) For the second quadratic equation \(p x^{2} + q x + r = 0\), the roots are \(\alpha+k\) and \(\beta+k\). By Vieta's formulas, we have: Sum of the roots: \((\alpha+k)+(\beta+k)=-\frac{q}{p}\) Product of the roots: \((\alpha+k)(\beta+k)=\frac{r}{p}\)

Substitute and simplify the relationships

From step 1, we substitute \(\alpha + \beta = -\frac{b}{a}\) into the sum of roots for the second equation: \((-\frac{b}{a}+2k) = -\frac{q}{p}\) Rearrange and solve for k: \(k=\frac{1}{2}\left(\frac{b}{a}-\frac{q}{p}\right)\)

Match the expression for k with the given options

Compare the expression for k obtained in step 2 with the given options: \(k=\frac{1}{2}\left(\frac{b}{a}-\frac{q}{p}\right)\) This expression matches option (c), so the correct answer is: \(k=\frac{1}{2}\left(\frac{b}{a}-\frac{q}{p}\right)\)

Key concepts

Vieta's Formulas

Understanding Vieta's formulas is crucial for anyone studying quadratic equations. These formulas relate the coefficients of a polynomial to sums and products of its roots. In the case of a quadratic equation of the form
\(ax^2 + bx + c = 0\), Vieta’s formulas tell us two key pieces of information:

• The sum of the roots \(\alpha\) and \(\beta\) is equal to the negative coefficient of the x term divided by the coefficient of the x^2 term, or \(-\frac{b}{a}\).
• The product of the roots \(\alpha\) and \(\beta\) is equal to the constant term divided by the coefficient of the x^2 term, or \(\frac{c}{a}\).

These relationships are immensely helpful when solving quadratic equations because they allow us to find the roots without necessarily having to perform the quadratic formula or factoring the polynomial.

Roots of a Quadratic Equation

The roots of a quadratic equation are the values of x that make the equation true (where the graph of the equation crosses the x-axis). For the general form \(ax^2 + bx + c = 0\), roots are commonly found using the quadratic formula, factoring, completing the square, or graphing.
However, the problem at hand provides a direct application of the roots through Vieta’s formulas. If we know the coefficients \(a\), \(b\), and \(c\), we can immediately tell the sum and product of the roots. This information can be invaluable, especially when comparing relationships between different quadratic equations, as seen in the given exercise where the roots of one equation are directly related to the roots of another by a constant \(k\).

Quadratic Relationships

Quadratic relationships go beyond the individual roots; they explore how changes in the quadratic equation affect the roots. In the given exercise, we see a case where the roots of one equation are modified by adding a constant \(k\), producing a new set of roots for another quadratic equation.
What the exercise demonstrates is that the change in the sum of the roots, resulting from adding \(k\) to each root, reflects a shift in the linear coefficient in accordance with Vieta’s formulas. By focusing on the shift in the sum of the roots, we gain insight into the constant \(k\), which represents the horizontal shift of the quadratic function on the x-axis. This awareness of quadratic relationships is essential for understanding more complex algebraic manipulations and graph transformations.