Quadratic Equations And Graphs

In this chapter, we will be discussing the below-mentioned topics in detail:

• Solving quadratic equations in one variable by factorisation
   

Let’s understand the concept with the help of an example:

\(\begin{align} 3 \times 0 &= 0 \\ 0 \times (-7) &= 0\\ p \times q &= pq \end{align}\)
 

If \(p\) times \(q\) is equal to \(0\), in that case, either \(p\) is equal to \(0\) or \(q\) is equal to \(0\).

The product of \(p\) and \(q\) is zero, but it is not certain whether \(p\) or \(q\) is zero, so because of that, we have used the word "or", i.e. either \(p\) is \(0\) "or" \(q\) is \(0\).

Zero Product Principle

If we have two unknowns multiplied together and the products of two unknowns are zero, that means either one of the unknowns is zero.

If \(A\) and \(B\) are algebraic expressions such that \(AB = 0\), then \(A=0\) or \(B = 0\).

Let’s understand this with the help of some examples:

 

Question 1:

Solve the following quadratic equations.

1.  \(x (x + 3) = 0\)
2.  \(x (3x – 6) = 0\)

Solution:

1. \(x (x + 3) = 0\)

\(\begin{align} \text{Either} \; x=0 \quad \text{or} \quad x+3 &=0 \\ \\ x &=-3 \end{align}\)

Linear equations give us one value of \(x\), but in quadratic equations, generally and not always, we will get two values of \(x\).

2. \(x (3x – 6) = 0\)

\(\begin{align} \text{Either} \; x=0 \quad \text{or} \quad 3x-6 &=0 \\ \\ 3x &=6 \\ \\ x &=2 \end{align}\)

Question 2:

Solve the following quadratic equations.

1.  \(\begin{align} x^2 – 7x &= 0 \end{align}\)
2.  \(\begin{align} 6x^2 + 21x = 0 \end{align}\)

Solution:

1.     \(\begin{align} x^2 – 7x &= 0 \end{align}\)  

\(\begin{align} x (x-7) &= 0 \end{align} ​\)

\(\begin{align} \text{Either}\; x = 0, \quad \text{or} \quad x – 7 &= 0 \\ \\ x &= 7 \end{align}\)

2.  \(\begin{align} 6x^2 + 21x &= 0 \end{align}\)

\(\begin{align} 3x (2x + 7) &= 0 \end{align}\)

\(\begin{align} \text{Either}\; 3x &= 0,& \text{or} \qquad 2x + 7 &= 0 \\ \\ x &= 0     &        2x &= -7 \\ \\ && x &= -3\frac {1}{2} \end{align}\)

Question 3:

Solve the following quadratic equations.

1. \(\begin{align} x^2 + 6x + 9 = 0 \end{align}\)
2. \(\begin{align} x^2 – 10x +25 = 0 \end{align}\)
3. \(\begin{align} 4x^2 – 9 &= 0 \end{align}\)

Solution: 

1. \(\begin{align} x^2 + 6x + 9 = 0 \end{align}\)

We know that, \(\begin{align} (a + b)^2 &= a^2 + 2ab + b^2 \end{align}\)

\(\begin{align} (x)^2 + 2 (x)(3) + (3)^2 &= 0 \\ \\ (x + 3)^2 &= 0 \\ \\ x + 3 &= 0 \\ \\ x &= -3 \end{align}\)

2. \(\begin{align} x^2 – 10x +25 = 0 \end{align}\)

We know that, \(\begin{align} (a - b)^2 &= a^2 - 2ab + b^2 \end{align}\)

\( \begin{align} (x)^2 – 2(x)(5) + (5)^2 &= 0 \\ \\ (x – 5)^2 &= 0 \\ \\ x – 5 &= 0 \\ \\ x &= 5 \end{align}\)

3. \(\begin{align} 4x^2 – 9 &= 0 \end{align}\)

We know that,  \(\begin{align} a^2 – b^2 &= (a+b)(a-b) \end{align}\)

\(\begin{align} (2x)^2 – (3)^2 &= 0 \\ \\ (2x + 3) (2x - 3) &= 0 \\\\ \text{Either}\qquad 2x + 3 &= 0 &\text{or} && 2x – 3 &= 0 \\ \\ 2x &= -3  &\text{or} &&   2x &= 3 \\ \\ x &= -1.5   &\text{or} &&   x &= 1.5  \end{align}\)

Question 4:

Solve the following quadratic equations.

1. \(\begin{align} x^2 – 7x +12 = 0 \end{align}\)
2. \(-3x^2 – 17x + 6 = 0\)

Solution:

1. Using Factorisation Table,

Multiplying diagonally,

\(\begin{align*} x^2 - 7x + 12 &= 0 \\ \\ (\;x\;– \;4\;) (\;x\;–\;3\;) &= 0 \\ \\ \text{Either}\qquad x-4 &= 0 &\text{or} && x\;–\;3 &= 0 \\ \\ x &= 4   &\text{or} &&   x &= 3  \end{align*}\)

2. Using Factorisation Table,

Multiplying diagonally,

\(\begin{align} -3x^2 - 17x + 6 &= 0 \\ \\ (-3x + 1) (x + 6) &= 0 \\ \\ \text{Either}\qquad -3x + 1 &= 0 &\text{or} && x + 6 &= 0 \\ \\ -3x &= -1   &\text{or} &&   x &= -6 \\ \\ x &= \frac{1}{3}   &\text{or} &&   x &= -6  \end{align}\)