[fs-toc-h2]Solutions to a Quadratic

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Before we go into solving polynomials via graphing, understanding how to solve one algebraically is still quite important in terms of the SAT and future math classes. There are multiple ways to solve a quadratic. Though, the most important and common way is by factoring.

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"Why do I need to know how to factor if I can just use desmos?" While solving a given quadratic by factoring isn't neccesary anymiore, the actuall skill and understanding of how it works can be tested in a multiude of ways (not just solving quadratics).

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If you need a refresher, or you weren't taught, the video below does a great job teaching the concept

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Now if you're given a quadratic and are asked to find its solution, 9/10 times the easiest and quickest way is by graphing. Take a look at the example below.\

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\[ 0 = 3x^2 + 21x + 18 \]

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Now you can divide by three and factor, use the quadratic equation, or even complete the square, but by graphing the equation you can get to an answer much quicker.

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In any quadratic function, the solution to the equation is where the graph crosses the x-axis. These are known as the x-intercepts.

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Find where your parabola crosses the x-axis (as this is when y is zero). Desmos has taken it upon themselves to highlight the zeros for you too.

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In the example above, your solutions are -6 and -1.

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NOTE: when solving a quadratic by graphing, make sure the equation is equal to zero or put in terms of y. If it isnt, the graph of the equation will not be a parabola.

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[fs-toc-h2]Number of Solutions and the Discriminant - The most tested subtopic.

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A quadratic equation can have three outcomes: 2 real solutions, 1 real solution, or no real solutions. This correlates with the number of times the graph of a quadratic crosses the x-axis. Remember, wherever the function crosses the x-axis is a solution.

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The three quadratics plotted below show the different amounts of solutions a quadratic can have.

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[fs-toc-h4]1 solution

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Since the quadratic above only touches the x-axis at one point, there is only one real solution.

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[fs-toc-h4]2 solutions

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Here it crosses the x-axis twice, meaning there are two real solutions.

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[fs-toc-h4]No solution

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Finally, this graph never crosses the x-axis, meaning it has no real solutions.

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Now on the SAT there's almost a 100% chance that you will be tested on this concept. Sometimes you'll be lucky enough where they will provide you with a full quadratic and simply ask how many real solutions there are. In that case, you can graph it on Desmos and see how many times it crosses the x-axis. Refer to the question below.

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[fs-toc-h4]Example 1

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Here we can see that the quadratic doesn't cross at a single point, meaning it has zero real solutions, making answer choice D correct.

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On many official tests though, this might not happen. Instead, you’ll be given a question like the one seen below.

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[fs-toc-h4]Example 2

Instead of asking how many solutions there are, they already tell you. Now you have to find which value for C will allow the quadratic to satisfy that specific amount of solutions. This is done using the discriminant.

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The discriminant of the quadratic equation is:

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\[\mathbf{b^2 - 4ac}\]

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Where \( a \), \( b \), and \( c \) are the respective coefficients in standard form (\( ax^2 + bx + c \)).

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If you plug in your three numbers into the discriminant and your answer is:

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Greater than zero (positive) = 2 real solutions
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Zero = 1 real solution
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Less than zero (negative) = no real solutions

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Let's look at our first example graph and its equation:
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\[ x^2 + 2x + 1 \]
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If I wanted to find the number of solutions of this equation algebraically, I would plug into the discriminant formula and compute.

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\[ b^2 - 4ac \]

\[ (2)^2 - 4(1)(1) \]

\[ 4 - 4 \]

\[ 0 \]

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The discriminant of this quadratic is equal to zero, meaning it has one solution. This is verified by our graph above.

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Try to use the discriminant to solve the question.

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We know that for an equation to have one real solution, the discriminant must equal zero. So let's substitute all our values in and solve from there.

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\[ b^2 - 4ac = 0 \]

\[ 30^2 - 4(-9)(c) = 0 \]

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Now all we have to do is expand our equation and solve like any other linear problem.

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\[ 30^2 - 4(-9)(c) = 0 \]

\[ 900 + 36c = 0 \]

\[ 36c = -900 \]

\[ c = -25 \]

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So the value for \( C \) that would make the quadratic above have one real solution is -25, answer choice C

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SHORTCUT: While knowing and understanding the discriminant is necessary, for these multiple-choice questions ,in particular, you can use Desmos. If you input any equation into desmos that has a single constant (in this case C), an option to add a slider will pop up. 

You can then widen the slider's range and move it until the graph only touches one point, the corresponding number will then be your answer.

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For more on the discriminant, click here and here.

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[fs-toc-h2]Minimum and Maximum

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There are two possible ways to find the vertex (which is either the minimum or the maximum of a function): using a formula or using desmos

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The only way you can immediately identify the vertex from a quadratic equation is if its already given in vertex form :  a(x-h)2+k

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For any quadratic that is  vertex form, its min/max happens at (h,k). For example, take the function

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\[f(x) = -(x - 9)^2 + 12\]

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The vertex is (9, 12). Since the parabola on this function opens downwards (due to the negative out in front), we can say that the maximum occurs at (9,12).

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If you're given a quadratic in standard form, you can use a simple equation. While you can simply plot a quadratic into desmos to find the vertex, in some cases this method will be a bit quicker.  

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The x-value of the vertex in a quadratic equation is given by the formula

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\[x_{\text{vertex}} = \frac{-b}{2a}\]

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PLEASE REMEMBER, this formula gives you the x-value of the vertex, NOT the y-value. Let's try to find the full vertex using our shortcut

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\[y=x^2 + 6x + 19\]

\[\frac{b}{-2a} = \frac{6}{-2(1)}\]

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x-value of the vertex is the -3

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To Find the Y-value of the vertex, plug in the respective x value back into the quadratic.

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\[y=x^2 + 6x + 19\]

\[y = (-3)^2 + 6(-3) + 19\]

\[y=10\]

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If you are adamant on using desmos, plot the equation they give you and the vertex will be highlighted

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If a question asks for the y-value of the vertex, it may be quicker to use desmos and simply plot it. However, if it asks for the x-value of the vertex, using the formula may only take you a couple of seconds.

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[fs-toc-h2]Different Forms & Their Graphs

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This has been alluded to previously in the lesson, but a quadratic can be written in three different ways. 

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Standard form : ax^2+bx+c

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Factored form : a(x-p)(x-q)

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Vertex form : a(x-h)^2 + k

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IMPORTANT NOTE: if ‘a’ is negative in any of these equations, the parabola of the equation will open downwards.  A negative quadratic opens downwards, a positive one opens upwards

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What’s very important to know is what these different equations tell you about their graphs. 

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[fs-toc-h4]Standard form

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In any standard form quadratic, the y-intercept is given through the constant c. Take the equation \(y=2x^2+3x+4\). 

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When plugging in 0 for x, you end up with 

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\(2(0)^2+3(0)+4\)

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All the terms with an x term cancel out and all your left with is 4. Meaning that the y-intercept of this equation is 4

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In the equation x^2+6x+7, the c term is 7, meaning the y-intercept is 7.

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The question above is directly testing this concept. Here we’re given a quadratic equation in standard form, meaning that the only piece of information that is shown directly as a constant is the y-intercept (2). Making answer choice C correct.

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[fs-toc-h4]Factored Form

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In most cases factored form is the most ‘useful’ form, as it directly gives you the solution to any quadratic or polynomial, no matter the degree.

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For any equation in form 

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(x-p) (x-q)

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The solutions/x-intercepts are p and q. This is true no matter the amount of factors. In a factored cubic equation 

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(x-p)(x-q)(x-j)

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The solutions are p, q, and j.

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Take for example the equation

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

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Remember, the zeros of the quadratic are the ‘opposite’ of what show up in the parenthesis. Since our first factor is subtracting 2 from x, then our solution is positive 2. Our second factor is adding 3, making our second solution negative 3. 2 or -3 will is is because a solution to a quadratic is whenever y equals zero. Plugging in either 2 or -3 will in turn make the whole equation equal zero. 

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\(y = (2-2)(2+3)\)

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\(y = (0)(5)\)

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\(y = 0\)

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or 

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\(y = (-3-2)(-3+3)\)

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\(y = (-5)(0)\)

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\(y = 0\) 

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Let's take a look at a couple of SAT questions

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X-intercepts, zeros, and solutions all mean the same thing: where the parabola intersects the x-axis. 

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Again, what we just went over, the factored form of a quadratic is the form in which the solutions/x-intercepts are displayed directly in the equation.

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The first three are in all different forms, but the fourth is in our perfect factored form, meaning that answer choice D is correct.

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This question is just as simple. The solutions to this equation are 6 and -7 (the opposite of what's in the parentheses). The question asks us the sum, so simply add both numbers

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\(6 + (-0.7) = 5.3\)

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Making answer choice C the correct answer. 

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[fs-toc-h4]Vertex form

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Vertex form gives you exactly what you think: the vertex (aka the max or min)

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In vertex form, \(y = a(x-h)^2 + k\) the vertex is given as \((h, k)\) 

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Take for example the equation

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\(y=2(x+4)^2+3\)

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The x-value of the vertex is simply the opposite sign of whatever is in the parenthesis. Here it's positive 4, so our x-value will be -4

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The y-value of the vertex is k. Here our k value is 3, so the y-value of the vertex is 3, giving us a full vertex of (-4, 3). This is confirmed by the graph below.

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NOTE: the coefficient a, in this case 2, does NOT affect the vertex in any way. 

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Take a look at the 2 problems below.

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The question is asking us which of the following equations can give us the coordinates of the vertex, which basically means “which equation is in vertex form”. 

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A and B are both in factored form, while A gives us the solutions to the graph, it doesn't tell us the vertex.

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C kind of looks like vertex form, but if you look closer the parenthesis aren't squared, meaning it's not correct either

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Leaving us with D, the only answer choice that's given in vertex form, making that the right choice.

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If you recall from earlier, a negative ‘a’ value will cause the parabola to open downwards. This means that both A and C are incorrect.

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The vertex of any equation in vertex form is (h, k), again, where h is opposite to whatever is in the parenthesis

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In the equation above, in the parenthesis were given -b, meaning the x-value of the vertex will be b

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The y-value is whatever is being added to the end (k), Which is C.

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So the full vertex is (b, c) and the graph opens downwards since a is negative, meaning the correct answer is choice B.

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For more on vertex form, click here.

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Summary:

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Standard form gives you the y-intercept : \(\mathbf{ax^2 + bx + c}\)

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Factored form gives you the zeros/x-intercepts : \(\mathbf{a(x-p)(x-q)}\)

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Vertex form gives you the vertex : \(\mathbf{a(x-h)^2 + k}\)

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