QUADRATIC EQUATIONS-Notes

A quadratic equation in variable \(x\) is an equation of the form: \[ax^2+bx+c=0\] where,
\(a,\ b,\ c\) are real numbers, \(a\ne 0\).

Examples \[\begin{aligned}2x^2-3x=5=0,\\ x^2=0,\\ -3x^2+9x + 1=0\end{aligned}\]

Any equation of the form \(p(x) = 0\), where \(p(x)\) is a polynomial of degree 2, is a quadratic equation.
But when we write the terms of \(p(x)\) in descending order of their degrees, then we get the standard form of the equation.
That is, \[ax^2 + bx + c = 0,\ a \ne 0\] is called the standard form of a quadratic equation.

Quadratic equations naturally appear in many situations:

• Area and Geometry Problems
  Examples includes:
  • Finding the sides of a rectangle when area and relation between sides are given.
  • Determining the diagonal of a square or dimensions of geometric shapes.
• Number Problems
  
  • Find two numbers whose sum is S and product is P
  • If a number is increased by its reciprocal, result is K
• Motion and Time Problems
  
  • Speed-distance relations involving squared terms.
  • Height of an object thrown upward.
• Practical Value Problems
  
  • Profit/Loss problems where cost and selling price appear in quadratic form.
  • Revenue maximization.

Method of Factorization

Factorization is used when the quadratic can be broken into two linear factors.

General Idea:

Convert \[ax^2+bx+c=0\] into \[(ax+m)(x+n)=0\] and then set each factor equal to zero.

Steps:

• Multiply \(a\times c\)
• Split the middle term \(b\) into two numbers whose product is \(ac\).
• Group and factor.
• Use Zero-Product Property: \[\text{if } \\pq=0\\ \implies p=0 \text{ or } q=0\]

For Example:

\[ \begin{aligned} x^2+7x+10&=0\\ x^2+5x+2x+10&=0\\ x(x+5)+2(x+5)&=0\\ (x+2)(x+5)&=0\\ \implies \text{either } x+2&=0,\text{ or}\\ x+5&=0\\ \implies x=-2 \text{ or } x&=-5 \end{aligned} \]

Completing the Square

This method transforms the quadratic into a perfect square.

Idea

Rewrite \[ax^2+bx+c=0\] as \[(x+p)^2=q\]

Steps

• Ensure coefficient of \(x^2\) is 1.
• Take half the coefficient of \(x\), square it, and add to both sides.
• Express left-hand side as a perfect square.
• Take square roots on both sides.
• Solve for \(x\)

This method is also the foundation for deriving the Quadratic Formula.

Quadratic Formula

This is the most universal method and works for every quadratic equation.
For Equation \[ax^2+bx+c=0\] the solutions are \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Important Term: Discriminant (D)

\[D=b^2-4ac\] The discriminant tells us the nature of the roots.

• \(D\gt 0\): Roots are real and distinct
  
  if \(D\gt 0\) then \[x_1,\ x_2 = \frac{-b\pm\sqrt{D}}{2a}\]
• \(D=0\): Real and equal root
  
  if \(D = 0\) then \[x = \frac{-b}{2a}\]
• \(D\lt 0\): No real roots
  
  if \(D\lt 0\) then
  
  the quadratic has no real solution (roots are complex).

Check whether the following are quadratic equations:

• \((x – 2)^2 + 1 = 2x – 3\)
• \((x + 2)^3 = x^3 – 4\)

Solution

$$\begin{aligned}\left( x-2\right) ^{2}+1=2x-3\\ x^{2}+4-4x+1-2x+3=0\\ x^{2}-6x+8=0\end{aligned}$$ is in the standard form of Quadratic Equation $$\begin{aligned}ax^{2}+bx+c=0\\ \therefore \left( x-2\right) ^{2}+1=2x-3 \\\text{is a Quadratic Equation}\end{aligned}$$

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$$\scriptsize\begin{aligned} (x+2) ^{3}=x^{3}-4\\ x^{3}+2^{3}+3x\cdot 2^{2}+3\cdot x^{2}.2-x^{3}+{4}=0\\ x^{3}+8+12x+6x^{2}-x^{3}+4=0\\ \Rightarrow 6x^{2}+12x+12=0\end{aligned}$$ is in the standard form of Quadratic Equation $$\begin{aligned}ax^{2}+bx+c=0\\ \therefore \left( x+2\right) ^{3}=x^{3}-4\\ \text{is a Quadratic Equation}\end{aligned}$$

Find the roots of the equation \(2x^2 – 5x + 3 = 0\), by factorisation.

Solution

$$2x^{2}-5x+3=0$$ By split middle term method, find two factors of \(ac\) such that their algebraic sum is 5 $$\begin{aligned}2\times 3\\ \text{sum }\Rightarrow 2+3=5\end{aligned}$$ $$\begin{aligned}2x^{2}-2x-3x+3=0\\ 2x\left( x-1\right) -3\left( x-1\right) =0\\ \left( x-1\right) \left( 2x-3\right) =0\\ \Rightarrow \text{either }(x-1) =0\\ \text{or } \left( 2x-3\right) =0\\ \Rightarrow x=1\\ x=\dfrac{3}{2}\end{aligned}$$ Roots are \(1\) and \(\frac{3}{2}\)

Find the roots of the quadratic equation \(6x^2 – x – 2 = 0\).

Solution

$$6x^{2}-x-2=0$$ Find the split middle term b factorising \(ac\) such that algebraic sum of factors is 1 $$\begin{aligned}6\times 2\\ 4\times 3\\ 4-3=1\end{aligned}$$ $$\begin{aligned}6x^{2}-4x+3x-2=0\\ \Rightarrow 6x^{2}+3x-4x-2=0\\ 3x\left( 2x+1\right) -2\left( 2x+1\right) =0\\ \left( 2x+1\right) \left( 3x-2\right) =0\end{aligned}$$ ⇒ Either $$\begin{aligned}2x+1=0\\ x=-\dfrac{1}{2}\end{aligned}$$ or $$\begin{aligned}3x-2=0\\ x=\dfrac{2}{3}\end{aligned}$$ therefore roots are $$\text{Roots are }\boxed{\dfrac{1}{2},\dfrac{2}{3}}$$

Find the roots of the quadratic equation \(3x^2-2\sqrt{6}x+2\) .

Solution

$$\scriptsize\begin{aligned}3x^{2}-2\sqrt{6}x+2=0\\ \left( \sqrt{3}x\right) ^{2}-2\sqrt{3}\cdot \sqrt{2}x+\left( \sqrt{2}\right) ^{2}=0\\ a^{2}-2ab+b^{2}\\ =\left( a-b\right) \left( a-b\right) \\ \left( \sqrt{3}x-\sqrt{2}\right) \left( \sqrt{3}x-\sqrt{2}\right) =0\\ \sqrt{3}x-\sqrt{2}=0\\ \sqrt{3}x=\sqrt{2}\\ x=\dfrac{\sqrt{2}}{\sqrt{3}}\\ x=\sqrt{\dfrac{2}{3}}\end{aligned}$$

Roots are \(\sqrt{\dfrac{2}{3}}\), \(\sqrt{\dfrac{2}{3}}\)

Find the discriminant of the quadratic equation \(2x^2 – 4x + 3 = 0\), and hence find the nature of its roots.

Solution

$$2x^{2}-4x+3=0$$ Discriminant \(D\) $$\begin{aligned}D&=b^{2}-4ac\\ &=\left( -4\right) ^{2}-4\left( 2\times 3\right)\\ &=16-24\\ &=-8\end{aligned}$$ \(D\lt 0\), Hence, Equation has no real roots

Find the discriminant of the equation \(3x^2 – 2x + \frac{1}{3}=0\) and hence find the nature of its roots. Find them, if they are real.

Solution

$$3x^{2}-2x+\dfrac{1}{3}=0$$

Multiplying both side of equation by 3

$$9x^{2}-6x+1=0$$

Discriminant

$$\begin{aligned}D&=b^{2}-4ac\\ &=\left( -6\right) ^{2}-4\left( 9\times 1\right) \\ &=36-36\\ &=0\end{aligned}$$

\(D=0\), Hence, Equation has equal root

Finding the roots

$$\begin{aligned} \text{Roots }&=\dfrac{-b}{2a}\\ &=\dfrac{-\left( -6\right) }{2\times 9}\\ &=\dfrac{1}{3},\dfrac{1}{3}\end{aligned}$$