For a real constant k, the quadratic equation x^2 - 2(k - 2)x + (k^2 - 3k + 2) = 0 has two… — ACT

For a real constant $k$ , the quadratic equation $x^2 - 2(k - 2)x + (k^2 - 3k + 2) = 0$ has two real roots, $r_1$ and $r_2$ . What is the value of $k$ that minimizes the sum of the squares of these roots, $r_1^2 + r_2^2$ ?

Answer: 2

Answer

The value of

k

that minimizes the sum of the squares of the roots is 2.

Applying Vieta's formulas and algebraic identities, the sum of the squares of the roots is expressed as

2k^2 - 10k + 12

. The discriminant condition for the roots to be real requires

k \leq 2

. Since the vertex of the upward-opening parabola

2k^2 - 10k + 12

is at

k = 2.5

, the function is strictly decreasing for all

k \leq 2

. Thus, the minimum value on the interval

k \leq 2

occurs at the boundary

k = 2

Step-by-Step Solution

Use Vieta's formulas to find the sum and product of the roots.

r_1 + r_2 = 2(k - 2)

and

r_1 r_2 = k^2 - 3k + 2

Vieta's formulas relate the roots of a quadratic equation

Ax^2 + Bx + C = 0

to its coefficients by

r_1 + r_2 = -B/A

and

r_1 r_2 = C/A

Express the sum of the squares of the roots,

r_1^2 + r_2^2

, in terms of

k

r_1^2 + r_2^2 = 2k^2 - 10k + 12

Since

r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1 r_2

, substituting the Vieta relations gives

[2(k-2)]^2 - 2(k^2 - 3k + 2) = 4(k^2 - 4k + 4) - 2k^2 + 6k - 4 = 2k^2 - 10k + 12

Determine the condition for the quadratic equation to have real roots using the discriminant.

\Delta = -4k + 8 \geq 0 \implies k \leq 2

For the roots

r_1

and

r_2

to be real, the discriminant

\Delta = B^2 - 4AC

must be greater than or equal to 0.

Minimize the quadratic function

f(k) = 2k^2 - 10k + 12

subject to the constraint

k \leq 2

k = 2

The parabola

f(k)

opens upwards with its vertex at

k = 2.5

. For

k \leq 2

, the function is strictly decreasing, meaning its minimum value on this interval occurs at the upper boundary,

k = 2

Key Concept

Quadratic Equations and the Quadratic Formula

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