Question

1. $x^2+5x-36=0$
2. $x^2+61=1-17x$
3. $3x+112=2x^2+22x$
4. $2x^2+7x=42-x$
5. $12x^2+17x=5$
6. $25x^2=20x-4$
7. $x^2+5x=0$
8. $8x^2+30x=2x$
9. $x^2=100$
10. $9x^2-6=10$
11. $x^2+5x-36=0$

$()-4)=0$

Explanation:

Problem 3: $3x^2+5x-36=0$

Step1: Identify coefficients for quadratic formula

$a=3$, $b=5$, $c=-36$

Step2: Apply quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$$ x=\frac{-5\pm\sqrt{5^2-4(3)(-36)}}{2(3)}=\frac{-5\pm\sqrt{25+432}}{6}=\frac{-5\pm\sqrt{457}}{6} $$

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Problem 4: $x^2+61=1-17x$

Step1: Rearrange to standard form

$x^2+17x+60=0$

Step2: Factor the quadratic

$(x+5)(x+12)=0$

Step3: Solve for $x$

$x+5=0 \implies x=-5$; $x+12=0 \implies x=-12$

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Problem 5: $3x+112=2x^2+22x$

Step1: Rearrange to standard form

$2x^2+19x-112=0$

Step2: Identify coefficients for quadratic formula

$a=2$, $b=19$, $c=-112$

Step3: Apply quadratic formula

$$ x=\frac{-19\pm\sqrt{19^2-4(2)(-112)}}{2(2)}=\frac{-19\pm\sqrt{361+896}}{4}=\frac{-19\pm\sqrt{1257}}{4} $$

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Problem 6: $2x^2+7x=42-x$

Step1: Rearrange to standard form

$2x^2+8x-42=0$

Step2: Simplify the equation

$x^2+4x-21=0$

Step3: Factor the quadratic

$(x+7)(x-3)=0$

Step4: Solve for $x$

$x+7=0 \implies x=-7$; $x-3=0 \implies x=3$

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Problem 7: $12x^2+17x=5$

Step1: Rearrange to standard form

$12x^2+17x-5=0$

Step2: Factor the quadratic

$(4x+5)(3x-1)=0$

Step3: Solve for $x$

$4x+5=0 \implies x=-\frac{5}{4}$; $3x-1=0 \implies x=\frac{1}{3}$

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Problem 8: $25x^2=20x-4$

Step1: Rearrange to standard form

$25x^2-20x+4=0$

Step2: Factor the quadratic

$(5x-2)^2=0$

Step3: Solve for $x$

$5x-2=0 \implies x=\frac{2}{5}$ (repeated root)

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Problem 9: $x^2+5x=0$

Step1: Factor out common term $x$

$x(x+5)=0$

Step2: Solve for $x$

$x=0$; $x+5=0 \implies x=-5$

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Problem 10: $8x^2+30x=2x$

Step1: Rearrange to standard form

$8x^2+28x=0$

Step2: Factor out common term $4x$

$4x(2x+7)=0$

Step3: Solve for $x$

$4x=0 \implies x=0$; $2x+7=0 \implies x=-\frac{7}{2}$

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Problem 11: $x^2=100$

Step1: Take square root of both sides

$x=\pm\sqrt{100}$

Step2: Simplify the root

$x=\pm10$

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Problem 12: $9x^2-6=10$

Step1: Isolate $x^2$ term

$9x^2=16 \implies x^2=\frac{16}{9}$

Step2: Take square root of both sides

$x=\pm\sqrt{\frac{16}{9}}$

Step3: Simplify the root

$x=\pm\frac{4}{3}$

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Answer:

1. $x=\frac{-5+\sqrt{457}}{6}$ and $x=\frac{-5-\sqrt{457}}{6}$
2. $x=-5$ and $x=-12$
3. $x=\frac{-19+\sqrt{1257}}{4}$ and $x=\frac{-19-\sqrt{1257}}{4}$
4. $x=-7$ and $x=3$
5. $x=-\frac{5}{4}$ and $x=\frac{1}{3}$
6. $x=\frac{2}{5}$ (repeated root)
7. $x=0$ and $x=-5$
8. $x=0$ and $x=-\frac{7}{2}$
9. $x=10$ and $x=-10$
10. $x=\frac{4}{3}$ and $x=-\frac{4}{3}$