Question

solve the following. check for extraneous solutions.

1. $sqrt{n}=8$
2. $sqrt3{x-5}=3$
3. $4sqrt{x-9}+2=14$
4. $sqrt{4x}-5=3$
5. $sqrt4{x+5}=2$
6. $sqrt{x+5}+3=1$
7. $x-6=sqrt{8x}$
8. $sqrt{3x+5}=sqrt{x-11}$

sat practice:
solve the system of equations using the elimination method.
$2x-3y=12$
$x+2y=-1$

Explanation:

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Problem 1: $\sqrt{n}=8$

Step1: Square both sides

$(\sqrt{n})^2=8^2$

Step2: Simplify to solve for $n$

$n=64$

Step3: Check for extraneous solutions

$\sqrt{64}=8$, which matches the original equation.

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Problem 2: $\sqrt{x-5}=3$

Step1: Square both sides

$(\sqrt{x-5})^2=3^2$

Step2: Simplify and solve for $x$

$x-5=9 \implies x=14$

Step3: Check for extraneous solutions

$\sqrt{14-5}=\sqrt{9}=3$, which matches the original equation.

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Problem 3: $4\sqrt{x-9}+2=14$

Step1: Isolate the radical term

$4\sqrt{x-9}=14-2=12$

Step2: Divide both sides by 4

$\sqrt{x-9}=\frac{12}{4}=3$

Step3: Square both sides

$(\sqrt{x-9})^2=3^2$

Step4: Simplify and solve for $x$

$x-9=9 \implies x=18$

Step5: Check for extraneous solutions

$4\sqrt{18-9}+2=4*3+2=14$, which matches the original equation.

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Problem 4: $\sqrt{4x}-5=3$

Step1: Isolate the radical term

$\sqrt{4x}=3+5=8$

Step2: Square both sides

$(\sqrt{4x})^2=8^2$

Step3: Simplify and solve for $x$

$4x=64 \implies x=\frac{64}{4}=16$

Step4: Check for extraneous solutions

$\sqrt{4*16}-5=\sqrt{64}-5=8-5=3$, which matches the original equation.

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Problem 5: $\sqrt[4]{x+5}=2$

Step1: Raise both sides to 4th power

$(\sqrt[4]{x+5})^4=2^4$

Step2: Simplify and solve for $x$

$x+5=16 \implies x=11$

Step3: Check for extraneous solutions

$\sqrt[4]{11+5}=\sqrt[4]{16}=2$, which matches the original equation.

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Problem 6: $\sqrt{x+5}+3=1$

Step1: Isolate the radical term

$\sqrt{x+5}=1-3=-2$

Step2: Analyze validity

A square root cannot equal a negative number, so there is no real solution.

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Problem 7: $x-6=\sqrt{8x}$

Step1: Square both sides

$(x-6)^2=(\sqrt{8x})^2$

Step2: Expand and simplify

$x^2-12x+36=8x \implies x^2-20x+36=0$

Step3: Factor the quadratic

$(x-2)(x-18)=0$

Step4: Solve for $x$

$x=2$ or $x=18$

Step5: Check for extraneous solutions

• For $x=2$: $2-6=-4$, $\sqrt{8*2}=\sqrt{16}=4$. $-4

eq 4$, so $x=2$ is extraneous.

• For $x=18$: $18-6=12$, $\sqrt{8*18}=\sqrt{144}=12$, which matches the original equation.

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Problem 8: $\sqrt{3x+5}=\sqrt{x-11}$

Step1: Square both sides

$3x+5=x-11$

Step2: Solve for $x$

$3x-x=-11-5 \implies 2x=-16 \implies x=-8$

Step3: Check for extraneous solutions

• $\sqrt{3*(-8)+5}=\sqrt{-19}$ (not a real number), $\sqrt{-8-11}=\sqrt{-19}$ (not a real number). Since square roots of negative numbers are not real, there is no real solution.

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SAT Practice: System of Equations

Given:
$2x-3y=12$ --- (1)
$x+2y=-1$ --- (2)

Step1: Eliminate $x$: Multiply eq(2) by 2

$2x+4y=-2$ --- (3)

Step2: Subtract eq(1) from eq(3)

$(2x+4y)-(2x-3y)=-2-12$

Step3: Simplify to solve for $y$

$7y=-14 \implies y=-2$

Step4: Substitute $y=-2$ into eq(2)

$x+2*(-2)=-1$

Step5: Solve for $x$

$x-4=-1 \implies x=3$

Step6: Check solution

• Eq(1): $2*3-3*(-2)=6+6=12$, which matches.
• Eq(2): $3+2*(-2)=3-4=-1$, which matches.

Answer:

1. $n=64$
2. $x=14$
3. $x=18$
4. $x=16$
5. $x=11$
6. No real solution
7. $x=18$
8. No real solution

SAT Practice: $x=3$, $y=-2$