Problem Set: Vieta Application of Quadratics¶

“Mathematics is the language in which the universe is written,” Galileo once said. These problems challenge you to not only solve but also to interpret the deep relationships between the roots and coefficients of quadratic equations—a vital aspect of higher-level problem-solving.

Here are some challenging problem sets based on Vieta's formulas, designed to test your understanding of the relationships between the roots and coefficients of quadratic equations:

Problem Set: Vieta’s Formulas¶

Problem 1.¶

The roots of the quadratic equation \(x^2 + px + q = 0\) are \(\alpha\) and \(\beta\). If \(\alpha + \beta = 7\) and \(\alpha\beta = -18\), find the quadratic equation.

Problem 2.¶

Given that the sum of the roots of a quadratic equation is \(10\) and the product of the roots is \(21\), find the quadratic equation.

Problem 3.¶

If one root of the equation \(x^2 - 5x + 6 = 0\) is doubled, find the new quadratic equation with the modified root.

Problem 4.¶

The sum of the squares of the roots of the quadratic equation is \(40\) and the sum of the roots is \(10\). Determine the quadratic equation.

Problem 5.¶

Find the quadratic equation whose roots are the reciprocals of the roots of the equation \(2x^2 + 5x - 3 = 0\).

Problem 6.¶

If \(\alpha\) and \(\beta\) are the roots of the equation \(x^2 + 4x - 8 = 0\), find the quadratic equation whose roots are \(\alpha^2\) and \(\beta^2\).

Problem 7.¶

The sum of the cubes of the roots of the quadratic equation is \(30\), and the product of the roots is \(12\). Find the quadratic equation.

Problem 8.¶

For the quadratic equation \(x^2 + px + q = 0\), it is known that the sum of the roots is equal to their product. Find the quadratic equation if \(p + q = -2\).

Problem 9.¶

The roots of the equation \(x^2 + 2x + 3 = 0\) are increased by 1. Find the new quadratic equation.

Problem 10.¶

If the roots of the equation \(x^2 - 7x + 10 = 0\) are swapped and their reciprocals are taken, what is the resulting quadratic equation?

Problem 11.¶

The roots of the quadratic equation are such that their product is 5 more than their sum. Find the quadratic equation if the product of the roots is 14.

Problem 12.¶

For the equation \(x^2 - kx + 4 = 0\), if the square of the sum of the roots is equal to the sum of the squares of the roots, find the value of \(k\) and the corresponding quadratic equation.

Solutions Hints:¶

Problem 1.¶

Use Vieta’s formulas:

\[ \alpha + \beta = -p, \quad \alpha\beta = q \]

So, the equation is:

\[ x^2 - 7x - 18 = 0 \]

Problem 2.¶

Using the sum and product of the roots, the equation is:

\[ x^2 - 10x + 21 = 0 \]

Problem 3.¶

Let the roots of the original equation be \(\alpha\) and \(\beta\), where \(\alpha + \beta = 5\) and \(\alpha\beta = 6\). If one root is doubled, say \(\alpha = 2\alpha\), the new sum of the roots becomes \((2\alpha + \beta)\) and the new product is \((2\alpha \beta)\).

Problem 4.¶

Use the identity:

\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]

Solve for the coefficients and find the equation.

Problem 5.¶

For reciprocal roots, the equation becomes:

\[ x^2 - 5x + 2 = 0 \]

Problem 6.¶

For squared roots, use the identities:

\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \quad \text{and} \quad \alpha^2\beta^2 = (\alpha\beta)^2 \]

Solve to get the new quadratic equation.

Problem 7.¶

Use the identities involving sums of cubes:

\[ \alpha^3 + \beta^3 = (\alpha + \beta)((\alpha + \beta)^2 - 3\alpha\beta) \]

Solve for the coefficients.

Problem 8.¶

Given \(\alpha + \beta = \alpha\beta\), use Vieta's formulas and the condition \(p + q = -2\) to find the equation.

Problem 9.¶

Let the roots be \(\alpha + 1\) and \(\beta + 1\). Use Vieta’s formulas to find the new equation.

Problem 10.¶

For reciprocal and swapped roots, the equation becomes:

\[ x^2 - \frac{7}{10}x + 1 = 0 \]

Problem 11.¶

Given \(\alpha\beta = 5 + (\alpha + \beta)\), solve for the equation.

Problem 12.¶

Use the identity for the sum of squares:

\[ (\alpha + \beta)^2 = \alpha^2 + \beta^2 \]

Solve for \(k\).

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