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Addison solves the equation below by first squaring both sides of the equation.

2x-1=sqrt(8-x)
What extraneous solution does Addison obtain?

x=

Addison solves the equation below by first squaring both sides of the equation. $\newline$ $2 x-1=\sqrt{8-x}$ $\newline$ What extraneous solution does Addison obtain? $\newline$ $x=$

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Evaluate recursive formulas for sequences

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Q. Addison solves the equation below by first squaring both sides of the equation.

\newline

2 x-1=\sqrt{8-x}

\newline

What extraneous solution does Addison obtain?

\newline

x=

Write Equation: Write down the original equation. $\newline$ $2x - 1 = \sqrt{8 - x}$
Square Both Sides: Square both sides of the equation to eliminate the square root. $\newline$ $(2x - 1)^2 = (\sqrt{8 - x})^2$
Perform Squaring: Perform the squaring on both sides. $\newline$ $(2x - 1)(2x - 1) = 8 - x$
Expand Left Side: Expand the left side of the equation using the FOIL method (First, Outer, Inner, Last). $\newline$ $4x^2 - 2x - 2x + 1 = 8 - x$
Combine Like Terms: Combine like terms on the left side of the equation. $\newline$ $4x^2 - 4x + 1 = 8 - x$
Add $x$ to Both Sides: Add $x$ to both sides of the equation to bring all $x$ terms to one side. $\newline$ $4x^2 - 4x + x + 1 = 8$
Subtract $8$ : Combine like terms on the left side of the equation. $4x^2 - 3x + 1 = 8$
Factor or Use Quadratic Formula: Subtract $8$ from both sides of the equation to set the equation to zero. $\newline$ $4x^2 - 3x + 1 - 8 = 0$
Calculate Discriminant: Combine like terms on the left side of the equation. $4x^2 - 3x - 7 = 0$
Find Solutions: Factor the quadratic equation, if possible, or use the quadratic formula to find the solutions for $x$ . The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 4$ , $b = -3$ , and $c = -7$ .
Check Solutions: Calculate the discriminant ( $b^2 - 4ac$ ) to determine the nature of the roots. $\newline$ Discriminant = $(-3)^2 - 4(4)(-7)$ $\newline$ Discriminant = $9 + 112$ $\newline$ Discriminant = $121$
Extraneous Solution: Since the discriminant is positive, there are two real solutions. Calculate the solutions using the quadratic formula. $\newline$ $x = \frac{3 \pm \sqrt{121}}{8}$ $\newline$ $x = \frac{3 \pm 11}{8}$
Extraneous Solution: Since the discriminant is positive, there are two real solutions. Calculate the solutions using the quadratic formula. $\newline$ $x = \frac{3 \pm \sqrt{121}}{8}$ $\newline$ $x = \frac{3 \pm 11}{8}$ Find the two solutions. $\newline$ $x = \frac{(3 + 11)}{8}$ or $x = \frac{(3 - 11)}{8}$ $\newline$ $x = \frac{14}{8}$ or $x = \frac{-8}{8}$ $\newline$ $x = 1.75$ or $x = -1$
Extraneous Solution: Since the discriminant is positive, there are two real solutions. Calculate the solutions using the quadratic formula. $\newline$ $x = \frac{3 \pm \sqrt{121}}{8}$ $\newline$ $x = \frac{3 \pm 11}{8}$ Find the two solutions. $\newline$ $x = \frac{(3 + 11)}{8}$ or $x = \frac{(3 - 11)}{8}$ $\newline$ $x = \frac{14}{8}$ or $x = \frac{-8}{8}$ $\newline$ $x = 1.75$ or $x = -1$ Check both solutions in the original equation to see if any of them is extraneous. $\newline$ For $x = 1.75$ : $\newline$ $2(1.75) - 1 = \sqrt{8 - 1.75}$ $\newline$ $x = \frac{3 \pm 11}{8}$ $0$ $\newline$ $x = \frac{3 \pm 11}{8}$ $1$ (True) $\newline$ For $x = -1$ : $\newline$ $x = \frac{3 \pm 11}{8}$ $3$ $\newline$ $x = \frac{3 \pm 11}{8}$ $4$ $\newline$ $x = \frac{3 \pm 11}{8}$ $5$ (False)
Extraneous Solution: Since the discriminant is positive, there are two real solutions. Calculate the solutions using the quadratic formula. $\newline$ $x = \frac{3 \pm \sqrt{121}}{8}$ $\newline$ $x = \frac{3 \pm 11}{8}$ Find the two solutions. $\newline$ $x = \frac{3 + 11}{8}$ or $x = \frac{3 - 11}{8}$ $\newline$ $x = \frac{14}{8}$ or $x = \frac{-8}{8}$ $\newline$ $x = 1.75$ or $x = -1$ Check both solutions in the original equation to see if any of them is extraneous. $\newline$ For $x = 1.75$ : $\newline$ $2(1.75) - 1 = \sqrt{8 - 1.75}$ $\newline$ $x = \frac{3 \pm 11}{8}$ $0$ $\newline$ $x = \frac{3 \pm 11}{8}$ $1$ (True) $\newline$ For $x = -1$ : $\newline$ $x = \frac{3 \pm 11}{8}$ $3$ $\newline$ $x = \frac{3 \pm 11}{8}$ $4$ $\newline$ $x = \frac{3 \pm 11}{8}$ $5$ (False)The solution $x = -1$ does not satisfy the original equation, so it is the extraneous solution. $\newline$ $x = -1$

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