Solve the given quadratic equation using the square root property. (x−12)2−5=0\left(x-\frac12\right)^2-5=0(x−21)2−5=0

x = 1 2 + 5 , x = 1 2 − 5 x=\frac12+\sqrt5,x=\frac12-\sqrt5 x = 2 1 + 5 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> , x = 2 1 − 5 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z">

Solve the given quadratic equation using the square root property. ( x − 1 2 ) 2 − 5 = 0 \left(x-\frac12\right)^2-5=0 ( x − 2 1 ) 2 − 5 = 0

Here, I want to solve this quadratic equation using the square root property and I have x minus 1 half squared minus 5 is equal to 0. So let's go ahead and start with step 1 and solve using the square root property. So we first want to just isolate our squared expression. So the expression I have being squared here is this x minus 1 half squared. So in order to isolate that, I just need to move my 5 over to the other side, which I can do by adding 5 to both sides. So we'll cancel, leaving me with x minus onetwo squared is equal to 5. Okay. So step 1 is done. Moving on to step 2, which is to take our positive and negative square root. So let's go ahead and take the square root of both sides. It will, of course, cancel on that left side leaving me with x minus 1 half, and that's equal to plus or minus the square root of 5. Okay. So step 2 is done as well. Let's go ahead and move on to step 3, which is to solve for x. So So in order to get x by itself here, I just need to move this one half to the other side, which I can do by adding 1 half. So if I add 1 half to both sides, it cancels on that left side leaving me with just x, which is exactly what I wanted, and that's equal to 1 half. I'm gonna go ahead and put that in the front there, plus or minus the square root of 5. Okay. So I've solved for x, and I'm actually done here. So we can split this into our 2 different solutions. This would just be 1 half plus root 5 and 1 half minus root 5. These answers are correct as well. Remember we could take these answers and plug them back in just to check, but I'm going to leave it off here. Let me know if you have any questions.

Solve the given quadratic equation using the square root property. ( x − 1 2 ) 2 − 5 = 0 \left(x-\frac12\right)^2-5=0 ( x − 2 1 ) 2 − 5 = 0

x = 1 2 + 5 , x = 1 2 − 5 x=\frac12+\sqrt5,x=\frac12-\sqrt5 x = 2 1 + 5 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> , x = 2 1 − 5 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z">

x = 5 2 , x = − 5 2 x=\frac52,x=-\frac52 x = 2 5 , x = − 2 5

x = 5 2 , x = − 5 2 x=\frac{\sqrt5}{2},x=-\frac{\sqrt5}{2} x = 2 5 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> , x = − 2 5 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z">

x = 1 2 , x = − 1 2 x=\frac12,x=-\frac12 x = 2 1 , x = − 2 1

0. Review of College Algebra

Solving Quadratic Equations

Trigonometry

0. Review of College Algebra

Solving Quadratic Equations

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Multiple Choice

Solve the given quadratic equation using the square root property.
$\left(x-\frac12\right)^2-5=0$

$x=\frac12+\sqrt5,x=\frac12-\sqrt5$

$x=\frac52,x=-\frac52$

$x=\frac{\sqrt5}{2},x=-\frac{\sqrt5}{2}$

$x=\frac12,x=-\frac12$

Verified step by step guidance

Start by isolating the squared term in the equation. The given equation is $(x - \frac{1}{2})^2 - 5 = 0$. Add 5 to both sides to get $(x - \frac{1}{2})^2 = 5$.

Apply the square root property, which states that if $a^2 = b$, then $a = \pm \sqrt{b}$. Therefore, take the square root of both sides: $x - \frac{1}{2} = \pm \sqrt{5}$.

Solve for $x$ by adding $\frac{1}{2}$ to both sides of the equation. This gives $x = \frac{1}{2} + \sqrt{5}$ and $x = \frac{1}{2} - \sqrt{5}$.

These are the two possible solutions for $x$ based on the square root property applied to the original equation.

Verify the solutions by substituting them back into the original equation to ensure they satisfy $(x - \frac{1}{2})^2 - 5 = 0$.

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