Quadratics
More Than Just a Smiley Face! 😊
Introduction
1. Introduction
Hey there! Ready to level up your algebra game? You've met quadratics before, those familiar expressions that give us beautiful parabolas. But now, we're going to dive deeper and uncover their real power. In this chapter, we'll master some seriously cool techniques. We'll start by learning how to complete the square, a powerful method that gives you the vertex of any parabola and lets you sketch it. Then, we'll meet the discriminant, a tiny part of the quadratic formula that tells us a big story about the number of solutions an equation has, without even solving it! 🕵️♂️ After that, we'll tackle quadratic inequalities, figuring out where a parabola is above or below the x-axis. We'll also see what happens when a straight line meets a curve by solving linear-quadratic simultaneous equations. Finally, we'll get clever and learn how to spot 'hidden' quadratics in more complex equations, transforming them into problems you already know how to solve. Let's get started!
2. Expressing Quadratics in Completed Square Form
Alright, let's talk about 'Completing the Square'. It sounds intense, but honestly, it's just giving a standard quadratic, , a major glow-up. We're rewriting it into the form . Why bother? Because this new form instantly tells you the vertex, or turning point, of the parabola: the minimum or maximum point.
The magic trick is to focus on the and terms. For , take the coefficient of , halve it (), and square it (). Add and subtract this value, so the expression's value doesn't change, and the first three terms become the perfect square .
The magic trick is to focus on the and terms. For , take the coefficient of , halve it (), and square it (). Add and subtract this value, so the expression's value doesn't change, and the first three terms become the perfect square .
When , there's one extra step at the start: factor out of the first two terms only, , complete the square inside the bracket, and remember to multiply the subtracted square by when it leaves the bracket.
. Once you have it in the form , the vertex is right there at . Notice the sign flip for the x-coordinate! The line of symmetry is .
Minimum, maximum and range. If the curve is a U shape and is the minimum value, so for all real the range is . If it is an upside-down U and is the maximum value, so the range is . If the domain is restricted, check whether the vertex is inside it; if it is not, the least (or greatest) value is at the end of the domain.
Sketching the graph. (1) Shape: U if , upside-down U if . (2) Vertex from the completed-square form. (3) -intercept: put , giving . (4) -intercepts: solve ; there may be two, one or none. (5) Draw a smooth symmetric curve and label every point with its coordinates. A sketch is not a plot: the examiner looks for the right shape and correctly labelled vertex and intercepts.
Minimum, maximum and range. If the curve is a U shape and is the minimum value, so for all real the range is . If it is an upside-down U and is the maximum value, so the range is . If the domain is restricted, check whether the vertex is inside it; if it is not, the least (or greatest) value is at the end of the domain.
Sketching the graph. (1) Shape: U if , upside-down U if . (2) Vertex from the completed-square form. (3) -intercept: put , giving . (4) -intercepts: solve ; there may be two, one or none. (5) Draw a smooth symmetric curve and label every point with its coordinates. A sketch is not a plot: the examiner looks for the right shape and correctly labelled vertex and intercepts.
Worked example
Worked Example: Finding the Vertex by Completing the Square
Let's Solve This Thing 🚀
Express in the form . Hence, state the coordinates of the vertex of the parabola , and sketch it.
- 1First up, the coefficient of isn't 1. We need to handle that by factoring out the 2 from the first two terms only. We'll let the just chill on the outside for a moment.
- 2Now, focus inside the bracket: . Take the coefficient of (which is ), halve it to get , and square it to get . We'll now add and subtract this inside the main bracket.
- 3The first three terms inside the bracket, , are now a perfect square. This factorises neatly into .
- 4We need to get the out of the main bracket. To do that, we have to multiply it by the 2 on the outside. It's like it has to pay an exit fee. So, .
- 5Tidy up the constant terms at the end by combining them: . This is our final, polished completed square form.
- 6Now for the big reveal. We compare our result to . The vertex is : flip the sign of the number in the bracket ( becomes ) and take the number on the end as it is (). Since , the vertex is a minimum point.
- 7For the sketch: a U shape, minimum at , crossing the -axis at . The -intercepts come from , so , about and . Label all of these on the curve.
Answer
3. Using the Discriminant to Determine the Nature of Roots
Alright, let's talk about one of the coolest shortcuts in quadratics: the discriminant. You know the quadratic formula, ? Well, the part living inside the square root, , is the VIP. We call it the discriminant, and its job is to tell you the 'nature of the roots' without you having to solve the whole thing.
Here's the breakdown:
1. If : The discriminant is positive. This means you'll be square-rooting a positive number, and the sign will give you two different answers. So, you get two distinct real roots. The parabola will cross the x-axis at two separate points.
Here's the breakdown:
1. If : The discriminant is positive. This means you'll be square-rooting a positive number, and the sign will give you two different answers. So, you get two distinct real roots. The parabola will cross the x-axis at two separate points.
2. If : The discriminant is zero. The square root of zero is just zero, so the part of the formula vanishes. You're left with just one value, . This is called a repeated root. On a graph, the parabola just touches the x-axis at its vertex. This is the critical condition for tangency: if a line is tangent to a curve, they meet at exactly one point, meaning the discriminant of their combined equation is zero.
3. If : The discriminant is negative. You can't square root a negative number and get a real answer. This means there are no real roots. The parabola will float entirely above or below the x-axis, never touching it.
For 'find the value of ' questions: rearrange to , write , and (some may contain ), then set , or as the question needs and solve for . If itself contains , check the value that makes separately, because then the equation is not a quadratic at all.
Worked example
Worked Example: Finding Values for a Condition of Tangency
When a Line and a Curve Just Barely Touch 🤝
The line is tangent to the curve . Find the possible values of the constant .
- 1First up, let's unpack the word 'tangent'. It's a huge clue! It means the line and the curve meet at exactly one point. To find intersection points, we always set the equations equal to each other. This will create a new quadratic, which must have a repeated root.
- 2Now, we need to shuffle everything over to one side to get it into the classic format. Be careful when you're combining the terms with .
- 3Since there's only one root (tangency!), we know the discriminant must be zero. Let's identify our and from the equation we just made. Here, , , and .
- 4Time to substitute these values into the discriminant equation. Don't forget that squaring the negative term will make it positive.
- 5Let's simplify and solve this new equation for . It's a simple quadratic in terms of .
- 6Now, we just need to take the square root of both sides to find the possible values for . Remember the ! The answer can be left in surd form unless specified otherwise.
Answer
4. Solving Quadratic Inequalities
Alright, let's level up from solving regular quadratic equations. With inequalities, we're not just looking for a couple of specific answers; we're finding a whole range of values where the condition is true. You cannot solve these like a linear inequality.
The game plan is simple but you've got to be precise. First, get everything onto one side of the inequality, so you have something like or . Next, find the critical values: solve the related equation, , by factorising, completing the square or the formula. These critical values are the points where the quadratic graph crosses the x-axis, the boundaries of your 'winning zone'.
Now for the most important part: deciding which side of the boundaries is the solution. The most reliable way to do this is to sketch the parabola. Seriously, don't skip this step! First, check the term. Is the coefficient 'a' positive? Then you've got a U-shaped parabola. Is 'a' negative? It's an upside-down U.
The game plan is simple but you've got to be precise. First, get everything onto one side of the inequality, so you have something like or . Next, find the critical values: solve the related equation, , by factorising, completing the square or the formula. These critical values are the points where the quadratic graph crosses the x-axis, the boundaries of your 'winning zone'.
Now for the most important part: deciding which side of the boundaries is the solution. The most reliable way to do this is to sketch the parabola. Seriously, don't skip this step! First, check the term. Is the coefficient 'a' positive? Then you've got a U-shaped parabola. Is 'a' negative? It's an upside-down U.
Mark your critical values on the x-axis and draw the correct shape passing through them. If your inequality was , you want the parts of the graph that are above the x-axis. If it was , you're looking for the part below the x-axis. For a U shape, that means gives two separate regions shooting outwards (like or ) and gives a single range between the critical values (like ). With or the critical values themselves are included.
"For all real ". A quadratic is positive for all real when it is a U shape () that never meets the axis (), and negative for all real when and .
"For all real ". A quadratic is positive for all real when it is a U shape () that never meets the axis (), and negative for all real when and .
Worked example
Worked Example: Solving a Quadratic Inequality
Let's Crack This Code
Find the set of values of for which .
- 1First up, we need to find the 'critical values'. To do this, we pretend the inequality is an equation and solve it to find the x-intercepts.
- 2Let's factorise the quadratic to find the roots. We're looking for two numbers that multiply to and add to .
- 3Now, we sketch the graph of . Since the coefficient of is positive (it's 1), we know it's a U-shaped parabola that crosses the x-axis at and .Sketch: a U-shaped curve crossing the x-axis at and .
- 4The original inequality was . The '' part means we want to find where the graph is above the x-axis. Looking at our sketch, this happens in two separate regions: to the left of and to the right of .The curve is above the axis for and for .
- 5Finally, we write this down as our solution. Use 'or' because x can't be in both places at once. It's either in one zone or the other.
Answer
5. Solving Simultaneous Linear and Quadratic Equations
Alright, let's talk about simultaneous equations. You've seen them before with two linear equations, where you find the one spot two straight lines cross. But what happens when one of them decides to be extra and curve? That's where a linear-quadratic system comes in. The solutions are literally the points of intersection on a graph.
The game plan is all about substitution. Rearrange the linear equation to get one variable by itself ('y' or 'x', whichever is simpler), then substitute that expression into the quadratic equation. That leaves a single quadratic equation in one variable.
The game plan is all about substitution. Rearrange the linear equation to get one variable by itself ('y' or 'x', whichever is simpler), then substitute that expression into the quadratic equation. That leaves a single quadratic equation in one variable.
Once you've done the substitution, you'll have a standard quadratic equation to solve: factorise it or use the formula. Solving this gives you the x-coordinate(s) of the intersection points. But wait, don't forget their partners! Each x-value has a corresponding y-value. To find them, substitute your x-values back into the original linear equation (it's way easier), and pair each with its own . The discriminant of the combined quadratic tells you how the line and curve interact: two solutions (the line cuts the curve), one repeated solution (the line is a tangent), or no real solutions (they miss).
The same method works when the quadratic is a circle, like with , or contains an term, like with : make one letter the subject of the linear equation and substitute it everywhere.
Worked example
Worked Example: Intersection of a Line and a Parabola
Let's Find Where They Meet Up ✨
Find the coordinates of the points of intersection of the line and the curve .
- 1First up, we need to get one variable by itself in the linear equation so we can substitute. Rearranging is super easy. Let's just move the over.
- 2Now for the main event: substitution. We're going to replace the 'y' in the quadratic equation with the expression we just found (). This combines the two equations into one that only has 'x' in it.
- 3Time to tidy this up. Let's move everything to one side to get a classic quadratic equation in the form .
- 4Solve for ! This quadratic looks nice and clean. We can factorise it. Notice that it's a perfect square.
- 5We only got one (repeated) x-value, which means the line is a tangent to the curve (it touches at exactly one point). Now, don't ghost the y-value! Substitute back into our easy linear equation () to find its partner.
- 6And we're done! State the final answer as a coordinate pair. This is the single point where the line and the curve touch.
Answer
6. Equations Reducible to Quadratic Form
Alright, let's talk about equations that are basically catfishing you. They show up looking all complicated, but underneath it all, they're just a quadratic in some function of . The key is to spot that one term is the square of another. For example, in , ; in , ; in , the square of appears. This is our signal!
When you spot this, your secret weapon is substitution. We introduce a new variable, usually , to stand in for the smaller piece. So, for the first equation above, we'd say, 'Let '. Then becomes , and the equation becomes a familiar friend: .
When you spot this, your secret weapon is substitution. We introduce a new variable, usually , to stand in for the smaller piece. So, for the first equation above, we'd say, 'Let '. Then becomes , and the equation becomes a familiar friend: .
From here, you just solve for like any normal quadratic. But (and this is super important) you're not done yet! You found , but the original question wants . So, you have to substitute back. If you found , you go back to and solve , giving . Reject impossible values of before you go back: and cannot be negative. For , gives , so ; reject , and gives . For , gives , so , and each value gives two angles in because repeats every .
Worked example
Worked Example: Solving a Quartic Equation via Substitution
Unmasking the Impostor Equation 🕵️♀️
Solve the equation .
- 1Recognise the quadratic form. Notice that the power of the first term () is double the power of the second term (). This is our clue that we can use a substitution.
- 2Introduce a substitution. Let's use the variable to simplify things. We'll set equal to . Then we rewrite the entire equation using .
- 3Solve the new quadratic equation for . We can solve this by factorising. We're looking for two numbers that multiply to and add to . That's and .
- 4Substitute back to find . We found the values for , but we need to find . We use our original substitution, , for each value of we found. Both values of are positive, so neither is rejected. Don't forget the when you square root!
- 5State the final solutions. We've found four values for . List them all out clearly.
Answer
Practice this in the app
Unlock the full chapter: practice questions, flashcards, mock papers and notes, free.
Continue revising