In the quadratic equation 2x^2 - 12x + k = 0, k is a constant. If the sum of the squares o… — SAT

In the quadratic equation $2x^2 - 12x + k = 0$ , $k$ is a constant. If the sum of the squares of the solutions to the equation is $26$ , what is the value of $k$ ?

Cevap: 10

Cevap

The value of

k

10

The correct answer is

10

. By expressing the sum of the squares of the solutions as

x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2

, we can substitute the sum of the solutions (

-\frac{-12}{2} = 6

) and the product of the solutions (

\frac{k}{2}

) directly into the expression. This gives

26 = 36 - k

. Solving for

k

yields

k = 10

. Alternatively, solving the quadratic equation using the quadratic formula yields solutions

3 + \frac{\sqrt{144-8k}}{4}

and

3 - \frac{\sqrt{144-8k}}{4}

. Squaring these solutions and setting their sum equal to

26

simplifies to

18 + 2\left(\frac{144-8k}{16}\right) = 26

, which also solves to

k = 10

Adım Adım Çözüm

Find the sum and product of the solutions using the coefficients of the quadratic equation.

The sum of the solutions is

6

and the product of the solutions is

\frac{k}{2}

By Vieta's formulas, for any quadratic equation

ax^2 + bx + c = 0

with solutions

x_1

and

x_2

, the sum of the solutions is

x_1 + x_2 = -\frac{b}{a}

and the product of the solutions is

x_1 x_2 = \frac{c}{a}

Apply the algebraic identity to express the sum of the squares of the solutions in terms of their sum and product.

x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2

This identity allows us to substitute the sum and product of the solutions directly without solving for the individual solutions first.

Substitute the values and solve for

k

26 = 6^2 - 2\left(\frac{k}{2}\right) \implies 26 = 36 - k \implies k = 10

Substituting the given sum of squares (

26

), the sum of solutions (

6

), and the product of solutions (

\frac{k}{2}

) allows us to solve for the unknown constant

k

directly.

Anahtar Kavram

Using the relationship between the roots and coefficients of a quadratic equation (Vieta's formulas) in combination with algebraic identities to solve for unknown constants.

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