$x-1=(2-x)^{2}$ $\newline$ What are the solutions to the given equation? $\newline$ Choose $1$ answer: $\newline$ (A) $x=0$ $\newline$ (B) $x=\frac{5-\sqrt{5}}{2}$ and $\newline$ $x=\frac{5+\sqrt{5}}{2}$ $\newline$ (C) $x=\frac{-1-\sqrt{21}}{2}$ and $\newline$ $x=\frac{-1+\sqrt{21}}{2}$ $\newline$ (D) The equation has no real solutions.

Full solution

x-1=(2-x)^{2}

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What are the solutions to the given equation?

\newline

Choose

1

answer:

\newline

(A)

x=0

\newline

(B)

x=\frac{5-\sqrt{5}}{2}

and

\newline

x=\frac{5+\sqrt{5}}{2}

\newline

(C)

x=\frac{-1-\sqrt{21}}{2}

and

\newline

x=\frac{-1+\sqrt{21}}{2}

\newline

(D) The equation has no real solutions.

Expand equation: First, we need to expand the right side of the equation $(2-x)^2$ . $(2-x)^2 = (2-x)(2-x) = 4 - 4x + x^2$
Rewrite equation: Now, we rewrite the original equation with the expanded form: $x - 1 = 4 - 4x + x^2$
Arrange equation: Next, we bring all terms to one side to set the equation to zero and arrange it in descending order of powers of x: $\newline$ $x^2 - 4x + x - 4 + 1 = 0$ $\newline$ $x^2 - 3x - 3 = 0$
Quadratic formula: We now have a quadratic equation. To solve for $x$ , we can use the quadratic formula $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$ , where $a = 1$ , $b = -3$ , and $c = -3$ .
Solve using formula: Plugging the values into the quadratic formula: $\newline$ x = $[-(-3) \pm \sqrt{(-3)^2 - 4(1)(-3)}] / (2(1))$ $\newline$ x = $[3 \pm \sqrt{9 + 12}] / 2$ $\newline$ x = $[3 \pm \sqrt{21}] / 2$
Simplify solution: Simplifying the square root and the expression: $\newline$ $x = \frac{3 \pm \sqrt{21}}{2}$ $\newline$ So we have two solutions: $\newline$ $x = \frac{3 + \sqrt{21}}{2}$ $\newline$ $x = \frac{3 - \sqrt{21}}{2}$
Compare solutions: We compare our solutions with the given options: $\newline$ (A) $x=0$ (Incorrect) $\newline$ (B) $x=\frac{5-\sqrt{5}}{2}$ and $x=\frac{5+\sqrt{5}}{2}$ (Incorrect) $\newline$ (C) $x=\frac{-1-\sqrt{21}}{2}$ and $x=\frac{-1+\sqrt{21}}{2}$ (Incorrect, sign error in the constant term) $\newline$ (D) The equation has no real solutions. (Incorrect, we found real solutions)