What is the product of all real solutions to the equation (x^2 - 6x + 10)^2 - 2(x^2 - 6x +… — GMAT

What is the product of all real solutions to the equation

(x^2 - 6x + 10)^2 - 2(x^2 - 6x + 10) - 35 = 0?

Cevap: 3

The product of all real solutions to the equation is 3.

Substituting

u = x^2 - 6x + 10

converts the given equation into

u^2 - 2u - 35 = 0

, which factors as

(u - 7)(u + 5) = 0

. Substituting back yields two quadratic equations:

x^2 - 6x + 3 = 0

and

x^2 - 6x + 15 = 0

. Checking the discriminants reveals that

x^2 - 6x + 3 = 0

has

D = 24 > 0

, producing two real roots with product

\frac{c}{a} = 3

, whereas

x^2 - 6x + 15 = 0

has

D = -24 < 0

, producing no real roots. Therefore, the product of all real solutions is 3.

Substitute a dummy variable to simplify the equation structure.

Let

u = x^2 - 6x + 10

. The equation becomes

u^2 - 2u - 35 = 0

Recognizing the repeated quadratic expression allows transforming a fourth-degree equation into a standard quadratic form.

Factor the quadratic equation in terms of

u

(u - 7)(u + 5) = 0 \implies u = 7 \text{ or } u = -5

Finding the values of

u

establishes the intermediate equations for

x

Analyze the first case

u = 7

for real solutions.

x^2 - 6x + 3 = 0

. Discriminant

D_1 = 36 - 12 = 24 > 0

. Product of real roots is

\frac{3}{1} = 3

A positive discriminant guarantees two real roots, and Vieta's formulas give their product directly.

Analyze the second case

u = -5

for real solutions.

x^2 - 6x + 15 = 0

. Discriminant

D_2 = 36 - 60 = -24 < 0

. No real roots.

A negative discriminant indicates complex conjugate roots, which must be excluded when finding the product of real solutions.

Combine results to find the final product of all real solutions.

Product = 3.

Only the two roots from the first case are real, so their product is the product of all real solutions.

Solving disguised quadratics via algebraic substitution and testing discriminants to filter out non-real roots before applying Vieta's formulas.

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