For a constant k, the quadratic equation x^2 - 2kx + k^2 - 4k + 3 = 0 has two distinct rea… — SAT

For a constant $k$ , the quadratic equation $x^2 - 2kx + k^2 - 4k + 3 = 0$ has two distinct real solutions. If both solutions are positive, which of the following describes all possible values of $k$ ?

$k > 3$ or $\frac{3}{4} < k < 1$ Answer
B
$k > 3$
C
$k > \frac{3}{4}$
D
$k < 1$ or $k > 3$

Answer

k > 3

\frac{3}{4} < k < 1

The correct answer is the option that includes both the interval between

\frac{3}{4}

and

1

and the interval greater than

3

. To ensure the quadratic equation has two distinct real solutions, the discriminant must be positive, which yields

k > \frac{3}{4}

. To ensure both solutions are positive, their sum and product must both be positive. The sum

2k > 0

requires

k > 0

. The product

k^2 - 4k + 3 > 0

requires

k < 1

k > 3

. The intersection of all these inequalities is

\frac{3}{4} < k < 1

k > 3

Step-by-Step Solution

Determine the condition for two distinct real solutions using the discriminant.

k > \frac{3}{4}

For the quadratic equation

x^2 - 2kx + k^2 - 4k + 3 = 0

to have two distinct real solutions, its discriminant

\Delta

must be strictly greater than

0

. Here,

a = 1

b = -2k

, and

c = k^2 - 4k + 3

. Thus,

\Delta = (-2k)^2 - 4(1)(k^2 - 4k + 3) = 4k^2 - 4k^2 + 16k - 12 = 16k - 12

. Setting

\Delta > 0

gives

16k - 12 > 0

, which simplifies to

k > \frac{3}{4}

Set up conditions for both solutions to be positive using the sum and product of the roots.

2k > 0

and

k^2 - 4k + 3 > 0

If a quadratic equation has two positive solutions, their sum must be positive and their product must also be positive. By Vieta's formulas, the sum of the solutions is

-\frac{b}{a} = 2k

and the product of the solutions is

\frac{c}{a} = k^2 - 4k + 3

Solve the inequalities for the sum and product of the roots.

k > 0

and either

k < 1

k > 3

The sum condition

2k > 0

simplifies to

k > 0

. The product condition

k^2 - 4k + 3 > 0

factors as

(k-1)(k-3) > 0

, which is satisfied when

k < 1

k > 3

Find the intersection of all the derived inequalities.

\frac{3}{4} < k < 1

k > 3

We must find the values of

k

that satisfy all three conditions:

k > \frac{3}{4}

k > 0

, and (

k < 1

k > 3

). Combining

k > \frac{3}{4}

and

k > 0

yields

k > \frac{3}{4}

. Combining

k > \frac{3}{4}

with the constraint that

k

must be less than

1

or greater than

3

gives the combined solution of

\frac{3}{4} < k < 1

k > 3

Key Concept

Using the discriminant and Vieta's formulas to determine the signs of the roots of a quadratic equation dependent on a parameter.

Alternative Method

Alternatively, you can write the solutions using the quadratic formula:

x = \frac{2k \pm \sqrt{16k - 12}}{2} = k \pm \sqrt{4k - 3}

. For both roots to be positive and distinct, we first need the term under the radical to be positive, so

4k - 3 > 0 \implies k > \frac{3}{4}

. Then, we need the smaller root to be positive:

k - \sqrt{4k - 3} > 0 \implies k > \sqrt{4k - 3}

. Since

k > \frac{3}{4} > 0

, we can square both sides:

k^2 > 4k - 3 \implies k^2 - 4k + 3 > 0 \implies (k - 1)(k - 3) > 0

. This yields

k < 1

k > 3

. Combining this with

k > \frac{3}{4}

gives the same result:

\frac{3}{4} < k < 1

k > 3

Estimated Time:3m 0s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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