What is the sum of the squares of the two real solutions to the quadratic equation \frac{1… — ACT

What is the sum of the squares of the two real solutions to the quadratic equation $\frac{1}{2}x^2 - 3x + 2 = 0$ ?

Cevap: 28

The sum of the squares of the two real solutions is 28.

Multiplying the equation

\frac{1}{2}x^2 - 3x + 2 = 0

by 2 yields

x^2 - 6x + 4 = 0

. By Vieta's formulas, the sum of the solutions is

x_1 + x_2 = 6

and the product of the solutions is

x_1 x_2 = 4

. Using the algebraic identity

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

, the sum of the squares is

6^2 - 2(4) = 36 - 8 = 28

. Alternatively, using the quadratic formula on the simplified equation gives solutions of

3 + \sqrt{5}

and

3 - \sqrt{5}

. Squaring these values yields

14 + 6\sqrt{5}

and

14 - 6\sqrt{5}

, which sum to 28.

Multiply the quadratic equation by 2 to clear the fraction.

x^2 - 6x + 4 = 0

Simplifying the fractional coefficients makes the equation easier to analyze and solve.

Identify the sum and product of the roots using Vieta's formulas.

x_1 + x_2 = 6

and

x_1 x_2 = 4

For a quadratic equation

ax^2 + bx + c = 0

, the sum of the roots is

-b/a

and the product is

c/a

Express the sum of the squares of the roots using the algebraic identity.

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

This identity allows the direct calculation of the sum of squares without needing to find the individual roots.

Substitute the values of the sum and product into the identity.

6^2 - 2(4) = 36 - 8 = 28

Plugging in the sum of 6 and product of 4 yields the final numerical answer.

Vieta's formulas state that for a quadratic equation

ax^2 + bx + c = 0

, the sum of the roots is

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

and the product of the roots is

x_1 x_2 = \frac{c}{a}

. Symmetric functions of roots like

x_1^2 + x_2^2

can be expressed in terms of these values.

Find the roots of the equation directly using the quadratic formula. After simplifying to

x^2 - 6x + 4 = 0

, the roots are

x = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(4)}}{2} = 3 \pm \sqrt{5}

. Squaring both solutions gives

(3 + \sqrt{5})^2 = 14 + 6\sqrt{5}

and

(3 - \sqrt{5})^2 = 14 - 6\sqrt{5}

. Adding these squares together yields

(14 + 6\sqrt{5}) + (14 - 6\sqrt{5}) = 28

Tahmini Süre:1m 30s

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