Quadratic Equations — Notes

Introduction to Quadratic Equations

A quadratic equation is an equation that can be expressed in the standard form $a{x}^2+bx+c=0$, where a, b, and c are real numbers and, crucially, a ≠ 0.

• Polynomial: A polynomial is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. For example, $x^2+2x+1$ is a polynomial.
• Quadratic Polynomial: A polynomial of degree 2. The general form is $a{x}^2+bx+c$.
• Quadratic Equation: Formed by equating a quadratic polynomial to zero. This means $a{x}^2+bx+c=0$. The condition $a\mathrm{\ne }0$ is essential because if $a=0$, the $x^2$ term vanishes, and the equation becomes $bx+c=0$, which is a linear equation, not quadratic.

Real-Life Applications

Quadratic equations are not just theoretical constructs; they appear frequently when modelling real-world scenarios.

How they are used: When a problem involves a relationship where a quantity depends on the square of another quantity, it often leads to a quadratic equation.

Example: Building a Prayer Hall

Imagine a charity trust wants to build a prayer hall with a carpet area of 300 square metres. The problem states that the length of the hall should be one metre more than twice its breadth.

Let the breadth of the hall be $x$ metres. According to the problem, the length of the hall is $2x+1$ metres.

The area of a rectangle is given by Area = Length × Breadth. So, for the prayer hall: Area $=(2x+1)\times x$ square metres Area $=(2x^2+x)$ square metres

We are given that the carpet area is 300 square metres. Therefore:

To solve this, we rearrange it into the standard quadratic equation form $a{x}^2+bx+c=0$:

This equation helps us find the breadth ($x$) of the hall, and subsequently its length.

Historical Context

The study and solution of quadratic equations have a long history:

• Babylonians: Solved problems equivalent to finding two positive numbers with a given sum and product, which could be reduced to $x^2-px+q=0$.
• Euclid: Developed geometrical methods to find lengths that are solutions to quadratic equations.
• Indian Mathematicians: Brahmagupta provided an explicit formula to solve equations of the form $a{x}^2+bx=c$. Sridharacharya later derived the quadratic formula by completing the square.
• Al-Khwarizmi: Studied various types of quadratic equations.

These historical developments show that quadratic equations have been a subject of mathematical interest for centuries, leading to methods that are still used today.

Remember

The defining characteristic of a quadratic equation is that the highest power of the variable is 2, and the coefficient of this squared term ($a$) must be non-zero.

Exam-smart

Real-life scenarios involving areas, projectile motion, or optimization problems often translate into quadratic equations. You might be given a word problem describing a situation, and you'll need to identify the variables, set up the quadratic equation, and then solve it (as we will learn in later sections) to find the required quantities.

Quadratic Equations

A quadratic equation in the variable $x$ is an equation that can be written in the standard form:

where $a$, $b$, and $c$ are real numbers, and crucially, $a\mathrm{\ne }0$.

The term $a{x}^2$ is the quadratic term, $bx$ is the linear term, and $c$ is the constant term. The condition $a\mathrm{\ne }0$ is essential because if $a=0$, the equation would become $bx+c=0$, which is a linear equation, not a quadratic one.

Any equation where the highest power of the variable is 2 is a quadratic equation. The standard form is obtained by arranging the terms in descending order of their powers.

Example 1: Representing Situations Mathematically

Given: John and Jivanti have 45 marbles together. They each lose 5 marbles, and the product of the remaining marbles is 124.

To find: The number of marbles they had initially.

Solution: Let the number of marbles John had be $x$. Then, the number of marbles Jivanti had is $45-x$. After losing 5 marbles each: Number of marbles left with John = $x-5$. Number of marbles left with Jivanti = $(45-x)-5=40-x$. The product of marbles left = $(x-5)(40-x)$. We are given this product is 124. So, $(x-5)(40-x)=124$. Expand the left side: $40x-x^2-200+5x=124$. Combine like terms: $-x^2+45x-200=124$. Rearrange into standard form (move all terms to one side and set to 0): $-x^2+45x-200-124=0$. $-x^2+45x-324=0$. Multiply by -1 to make the leading coefficient positive: $x^2-45x+324=0$. This is the required quadratic equation.

Answer: The quadratic equation representing the situation is $x^2-45x+324=0$.

Example 2: Checking if an Equation is Quadratic

Given: The equation $(x-2{)}^2+1=2x-3$.

To find: Whether this is a quadratic equation.

Solution: Simplify the Left Hand Side (LHS): LHS = $(x-2{)}^2+1$. Using the identity $(a-b{)}^2=a^2-2ab+b^2$: LHS = $(x^2-2(x)(2)+2^2)+1$. LHS = $(x^2-4x+4)+1$. LHS = $x^2-4x+5$. Now, set the simplified LHS equal to the Right Hand Side (RHS): $x^2-4x+5=2x-3$. Rearrange into the standard form $a{x}^2+bx+c=0$: $x^2-4x-2x+5+3=0$. $x^2-6x+8=0$. This equation is in the form $a{x}^2+bx+c=0$ with $a=1$, $b=-6$, and $c=8$. Since $a\mathrm{\ne }0$, it is a quadratic equation.

Answer: The given equation is a quadratic equation.

Example 3: Identifying Non-Quadratic Equations

Given: The equation $x(x+1)+8=(x+2)(x-2)$.

To find: Whether this is a quadratic equation.

Solution: Simplify the LHS: LHS = $x(x+1)+8=x^2+x+8$. Simplify the RHS using the identity $(a+b)(a-b)=a^2-b^2$: RHS = $(x+2)(x-2)=x^2-2^2=x^2-4$. Now, set the simplified LHS equal to the RHS: $x^2+x+8=x^2-4$. Rearrange to check the form: $x^2-x^2+x+8+4=0$. $x+12=0$. This is a linear equation, as the $x^2$ terms cancelled out. It is not in the form $a{x}^2+bx+c=0$ with $a\mathrm{\ne }0$.

Answer: The given equation is not a quadratic equation.

Example 4: Quadratic Equations from Word Problems

Given: The area of a rectangular plot is 528 $m^2$. The length is one more than twice its breadth.

To find: The length and breadth of the plot, represented by a quadratic equation.

Solution: Let the breadth of the rectangular plot be $x$ metres. According to the problem, the length is one more than twice its breadth. So, Length = $2x+1$ metres. The area of a rectangle is given by Length $\times$ Breadth. Area = $(2x+1)\times x$. We are given that the Area = 528 $m^2$. So, $x(2x+1)=528$. Expand the LHS: $2x^2+x=528$. Rearrange into standard quadratic form: $2x^2+x-528=0$. This is the required quadratic equation.

Answer: The quadratic equation is $2x^2+x-528=0$.

Remember

A quadratic equation always has a non-zero coefficient for the $x^2$ term. Simplifying the equation is crucial to determine its true nature, as terms might cancel out.

Exam-smart

You might be given a complex-looking equation or a word problem and asked to determine if it's quadratic, or to write it in the standard form $a{x}^2+bx+c=0$. The key is algebraic manipulation to simplify and group terms correctly. For word problems, carefully identify the variable, express other quantities in terms of it, and use the given conditions to form the equation.

Solution of a Quadratic Equation by Factorisation

A root of a quadratic equation $a{x}^2+bx+c=0$ (where $a\mathrm{\ne }0$) is a real number $\alpha$ such that if we substitute $\alpha$ for $x$, the equation holds true: $a{\alpha }^2+b\alpha +c=0$. These roots are also called the solutions of the quadratic equation. A quadratic equation can have at most two roots.

The method of factorisation is used to find these roots by rewriting the quadratic polynomial as a product of two linear factors. This process relies on the Zero Product Property, which states that if the product of two or more factors is zero, then at least one of the factors must be zero.

How Factorisation Works for Quadratic Equations:

1. Splitting the Middle Term: The middle term ($bx$) of the quadratic equation $a{x}^2+bx+c=0$ is split into two terms, say $px$ and $qx$, such that $px+qx=bx$. The key condition for splitting is that the product of these two new terms, $px\times qx$, must be equal to the product of the first and last terms of the original quadratic: $(px)(qx)=(a{x}^2)(c)$.
2. Grouping and Factoring: After splitting the middle term, the polynomial is grouped into two pairs. Common factors are then taken out from each pair. The aim is to obtain a common binomial factor.
3. Applying Zero Product Property: The quadratic equation is now expressed as a product of two linear factors. By setting each linear factor equal to zero, we get two linear equations.
4. Solving Linear Equations: Solving these linear equations gives us the roots (solutions) of the original quadratic equation.

Example 1: Find the roots of the equation $2x^2-5x+3=0$ by factorisation.

Given: The quadratic equation $2x^2-5x+3=0$. To find: The roots of the equation.

Solution: First, we need to split the middle term, $-5x$. We look for two numbers whose product is $(2x^2)\times 3=6x^2$ and whose sum is $-5x$. These numbers are $-2x$ and $-3x$.

Now, we group the terms:

Factor out common terms from each group:

Notice the common binomial factor $(x-1)$:

So, the original equation becomes:

Now, apply the Zero Product Property: Either $2x-3=0$ or $x-1=0$. Solve each linear equation: If $2x-3=0$, then $2x=3$, so $x=\frac{3}{2}$. If $x-1=0$, then $x=1$. Answer: The roots are $x=1$ and $x=\frac{3}{2}$.

Example 2: Find the roots of the quadratic equation $6x^2-x-2=0$.

Given: The quadratic equation $6x^2-x-2=0$. To find: The roots of the equation.

Solution: We split the middle term, $-x$. We need two numbers whose product is $(6x^2)\times (-2)=-12x^2$ and whose sum is $-x$. These numbers are $3x$ and $-4x$.

Group the terms:

Factor out common terms:

Factor out the common binomial $(2x+1)$:

So, the equation becomes:

Apply the Zero Product Property: Either $3x-2=0$ or $2x+1=0$. Solve each linear equation: If $3x-2=0$, then $3x=2$, so $x=\frac{2}{3}$. If $2x+1=0$, then $2x=-1$, so $x=-\frac{1}{2}$. Answer: The roots are $x=\frac{2}{3}$ and $x=-\frac{1}{2}$.

Example 3: Find the roots of the quadratic equation $3x^2-6x+2=0$.

Given: The quadratic equation $3x^2-6x+2=0$. To find: The roots of the equation.

Solution: We need to split the middle term, $-6x$. We are looking for two numbers whose product is $(3x^2)\times 2=6x^2$ and whose sum is $-6x$. Let's list factors of $6x^2$: $(1,6),(2,3),(-1,-6),(-2,-3)$. Their sums are $7x,5x,-7x,-5x$. None of these sum to $-6x$. This indicates that the quadratic polynomial $3x^2-6x+2$ cannot be easily factorised into linear factors with integer coefficients by splitting the middle term. The provided textbook solution seems to have made an error in its factorisation steps, as $3x^2-2\sqrt{6}x+2=0$ has roots $\sqrt{2\mathrm{/}3}$, not what is shown. We will use a correct example of a repeated root from the exercise section.

Example 4: Find the roots of the quadratic equation $100x^2-20x+1=0$.

Given: The quadratic equation $100x^2-20x+1=0$. To find: The roots of the equation.

Solution: We need to split the middle term, $-20x$. We look for two numbers whose product is $(100x^2)\times 1=100x^2$ and whose sum is $-20x$. These numbers are $-10x$ and $-10x$.

Group the terms:

Factor out common terms:

Factor out the common binomial $(10x-1)$:

So, the equation becomes:

Apply the Zero Product Property:

Solve the linear equation: $10x=1$, so $x=\frac{1}{10}$. Since both factors are identical, the root is repeated. Answer: The roots are $x=\frac{1}{10}$ and $x=\frac{1}{10}$ (a repeated root).

Example 5: Find the dimensions of the prayer hall.

Given: A prayer hall with breadth $x$ metres. The equation for its breadth is $2x^2+x-300=0$. To find: The dimensions (breadth and length) of the prayer hall.

Solution: We need to find the roots of the quadratic equation $2x^2+x-300=0$ using factorisation. Split the middle term, $x$. We need two numbers whose product is $(2x^2)\times (-300)=-600x^2$ and whose sum is $x$. These numbers are $25x$ and $-24x$.

Group the terms:

Factor out common terms:

Factor out the common binomial $(2x+25)$:

So, the equation becomes:

Apply the Zero Product Property: Either $x-12=0$ or $2x+25=0$. Solve each linear equation: If $x-12=0$, then $x=12$. If $2x+25=0$, then $2x=-25$, so $x=-\frac{25}{2}$ or $-12.5$. Since $x$ represents the breadth of the hall, it must be a positive value. Therefore, the breadth $x=12$ metres. The length of the hall is given by $2x+1$. Length $=2(12)+1=24+1=25$ metres. Answer: The dimensions of the prayer hall are breadth 12 m and length 25 m.

Remember

When splitting the middle term $bx$ in $a{x}^2+bx+c=0$, the two numbers, say $p$ and $q$, must satisfy both $p+q=b$ and $p\times q=a\times c$.

Exam-smart

Case/source questions will often provide a word problem (like the prayer hall or train problems). You will need to first formulate the quadratic equation from the given information and then solve it using factorisation. Pay attention to the context of the problem to discard negative or physically impossible solutions (e.g., negative length, negative age, negative speed).

Try this: Find the roots of the quadratic equation $x^2-7x+10=0$ by factorisation. Answer: $x=5$ and $x=2$.

Nature of Roots

The nature of roots of a quadratic equation $a{x}^2+bx+c=0$ (where $a\mathrm{\ne }0$) refers to whether the roots are real or not, and if they are real, whether they are distinct or equal. This nature is determined by a specific expression called the discriminant.

The discriminant of a quadratic equation $a{x}^2+bx+c=0$ is given by the expression:

The value of the discriminant ($D$) tells us about the roots:

Understanding the Discriminant's Role

The quadratic formula for the roots of $a{x}^2+bx+c=0$ is:

Notice that the term under the square root is precisely the discriminant ($D=b^2-4ac$).

• If $D>0$: $\sqrt{D}$ is a positive real number. This gives us two different values for $x$: one with $+\sqrt{D}$ and one with $-\sqrt{D}$. Hence, two distinct real roots.
• If $D=0$: $\sqrt{D}=\sqrt{0}=0$. The formula becomes $x=\frac{-b\pm 0}{2a}$, which simplifies to $x=\frac{-b}{2a}$. Both possibilities of $\pm$ yield the same value, so we have two equal real roots.
• If $D<0$: $\sqrt{D}$ involves the square root of a negative number. In the set of real numbers, there is no number whose square is negative. Therefore, there are no real numbers that can be solutions to the equation; the roots are imaginary.

Example 7: Finding the Discriminant and Nature of Roots

Given: The quadratic equation $2x^2-4x+3=0$. To find: The discriminant and the nature of its roots.

Solution: The given equation is in the form $a{x}^2+bx+c=0$. Comparing, we get:

The discriminant is $D=b^2-4ac$. Substitute the values of $a$, $b$, and $c$:

Since $D=-8$, which is less than 0 ($D<0$), the quadratic equation has no real roots.

Answer: The discriminant is -8. The nature of roots is no real roots.

Example 9: Finding Equal Real Roots

Given: The quadratic equation $3x^2-2x+\frac{1}{3}=0$. To find: The discriminant, the nature of its roots, and the roots if they are real.

Solution: The given equation is in the form $a{x}^2+bx+c=0$. Comparing, we get:

The discriminant is $D=b^2-4ac$. Substitute the values of $a$, $b$, and $c$:

Since $D=0$, the quadratic equation has two equal real roots.

To find the roots, we use the quadratic formula $x=\frac{-b\pm \sqrt{D}}{2a}$:

This gives two identical roots:

Answer: The discriminant is 0. The nature of roots is two equal real roots. The roots are $\frac{1}{3}$ and $\frac{1}{3}$.

Application of Discriminant in Problems

The concept of the discriminant is crucial for determining the possibility of real solutions in various real-world scenarios that can be modeled by quadratic equations.

Example 8: Is it Possible to Erect a Pole?

Given: A circular park of diameter 13 metres. A pole is to be erected on the boundary such that the difference of its distances from two diametrically opposite gates A and B is 7 metres. To find: If it's possible to erect the pole, and if so, find the distances from the gates.

Solution: Let the circular park have center O. Let A and B be the two diametrically opposite gates on the boundary. The diameter AB = 13 m. Let P be the point on the boundary where the pole is to be erected. Let the distance of the pole from gate B be $x$ metres ($BP=x$). The difference of its distances from A and B is 7 metres. So, $AP-BP=7$ or $BP-AP=7$. If $AP-BP=7$, then $AP=BP+7=x+7$. If $BP-AP=7$, then $AP=BP-7=x-7$. We will consider $AP=x+7$ and $BP=x$ for now, as distances must be positive.

Since P is on the boundary and AB is the diameter, the angle subtended by the diameter at any point on the circumference is 90°. So, $\mathrm{\angle }APB={90}^{\circ }$. In right-angled triangle APB, by the Pythagorean theorem:

Divide the equation by 2:

To determine if it's possible to erect the pole, we check if this quadratic equation has real roots by calculating its discriminant:

Since $D=289>0$, the equation has two distinct real roots. Therefore, it is possible to erect the pole.

Now, we find the roots using the quadratic formula:

Two possible values for $x$:

Since $x$ represents a distance, it must be positive. So, we ignore $x=-12$. Thus, $x=5$ m.

The distances from the gates are: Distance from gate B ($BP$) = $x=5$ m. Distance from gate A ($AP$) = $x+7=5+7=12$ m.

Answer: Yes, it is possible. The pole should be erected at a distance of 5 m from gate B and 12 m from gate A.

Example from Exercise: Designing a Rectangular Mango Grove

Given: A rectangular mango grove with length twice its breadth, and an area of 800 m${}^2$. To find: If it's possible to design such a grove, and if so, find its length and breadth.

Solution: Let the breadth of the rectangular grove be $b$ metres. The length is twice the breadth, so length $l=2b$ metres. The area of the rectangle is given by $Area=length\times breadth$.

Rearrange into a quadratic equation:

Divide by 2:

This is a quadratic equation in $b$. We can solve it by factoring or using the discriminant. The form is $a{b}^2+bb+c=0$, with $a=1$, $b=0$ (coefficient of $b$), and $c=-400$.

Since $D=1600>0$, the equation has two distinct real roots for $b$. Therefore, it is possible to design such a grove.

Now, find the roots for $b$:

Two possible values for $b$:

Since breadth must be a positive value, we take $b=20$ m. The length $l=2b=2\times 20=40$ m.

Answer: Yes, it is possible. The breadth is 20 m and the length is 40 m.

Remember

When dealing with real-world problems (like lengths, ages, or quantities), if a quadratic equation yields both a positive and a negative solution, always choose the positive one because physical quantities cannot be negative.

Exam-smart

A question might ask about the "possibility" of a situation. This is a direct cue to calculate the discriminant. If $D\ge 0$, the situation is possible; if $D<0$, it's impossible. The actual values of the roots would then be calculated if the situation is possible.