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$trative-math //×6418b49dfbc9d0c :quadratic-quations-part2/x6418b49dfbc9d0c9:apply-quadratio-formula-part2... All Bookmarks Khan Academy Donate \square nicolel3754 Tori and Gavin were trying to solve the equation: (x+1)^(2)-3=13 Tori said, "I'Il add 3 to both sides of the equation and solve using square roots." Gavin said, "I'll multiply (x+1)^(2) and rewrite the equation as x^(2)+2x+1-3=13. Then I'll subtract 13 from both sides, combine like terms, and solve using the quadratic formula with a=1,b=2, and c=-15." Whose solution strategy would work? Choose 1 answer: (A) Only Tori's (8) Only Gavin's (c) Both$

trative-math $/ \times 6418 \mathrm{~b} 49 \mathrm{df} b \mathrm{c} 9 \mathrm{~d} 0 \mathrm{c}$ :quadratic-quations-part $2$ /x $6418$ b $49$ dfbc $9$ d $0$ c $9$ :apply-quadratio-formula-part $2$ ... $\newline$ All Bookmarks $\newline$ Khan Academy $\newline$ Donate \square $\newline$ nicolel $3754$ $\newline$ Tori and Gavin were trying to solve the equation: $\newline$ $(x+1)^{2}-3=13$ $\newline$ Tori said,

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Algebra 1

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Full solution

Q. trative-math

/ \times 6418 \mathrm{~b} 49 \mathrm{df} b \mathrm{c} 9 \mathrm{~d} 0 \mathrm{c}

:quadratic-quations-part

2

6418

49

dfbc

9

0

9

:apply-quadratio-formula-part

2

...

\newline

All Bookmarks

\newline

Khan Academy

\newline

Donate \square

\newline

nicolel

3754

\newline

Tori and Gavin were trying to solve the equation:

\newline

(x+1)^{2}-3=13

\newline

Tori said,

Start with Tori's method: Start with Tori's method. $\newline$ Equation: $(x+1)^2 - 3 = 13$ $\newline$ Add $3$ to both sides: $(x+1)^2 - 3 + 3 = 13 + 3$ $\newline$ Simplify: $(x+1)^2 = 16$
Solve using square roots: Solve using square roots. $\newline$ Take the square root of both sides: $x+1 = \pm \sqrt{16}$ $\newline$ Simplify: $x+1 = \pm 4$
Solve for x: Solve for $x$ . $\newline$ Case $1$ : $x+1 = 4$ $\newline$ Subtract $1$ from both sides: $x = 3$ $\newline$ Case $2$ : $x+1 = -4$ $\newline$ Subtract $1$ from both sides: $x = -5$
Check Gavin's method: Check Gavin's method. $\newline$ Rewrite the equation: $(x+1)^2 - 3 = 13$ $\newline$ Expand: $x^2 + 2x + 1 - 3 = 13$ $\newline$ Simplify: $x^2 + 2x - 2 = 13$ $\newline$ Subtract $13$ from both sides: $x^2 + 2x - 2 - 13 = 0$ $\newline$ Combine like terms: $x^2 + 2x - 15 = 0$
Solve using the quadratic formula: Solve using the quadratic formula. $\newline$ Quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $\newline$ Here, $a = 1$ , $b = 2$ , $c = -15$ $\newline$ Calculate discriminant: $b^2 - 4ac = 2^2 - 4(1)(-15) = 4 + 60 = 64$
Continue solving: Continue solving. $\newline$ Calculate roots: $x = \frac{-2 \pm \sqrt{64}}{2(1)} = \frac{-2 \pm 8}{2}$ $\newline$ Case $1$ : $x = \frac{-2 + 8}{2} = \frac{6}{2} = 3$ $\newline$ Case $2$ : $x = \frac{-2 - 8}{2} = \frac{-10}{2} = -5$
Conclusion: Conclusion. $\newline$ Both Tori and Gavin's methods give the same solutions: $x = 3$ and $x = -5$