Quadratic Equations Easy Guide

A quadratic equation is a second-degree polynomial equation in one variable, where the highest power of the variable is 2. General form of quadratic equation is 𝑎𝑥² + 𝑏𝑥 + 𝑐 = 0
where:

• a,b, and c are real numbers,
• a≠0 
• x is the variable.

if a=0, the equation becomes linear instead of quadratic and if  b=0 the equation becomes a pure quadratic equation.
A pure quadratic equation is a special case where the linear term (𝑏𝑥) is missing: 𝑎𝑥² + 𝑐 = 0 where 𝑏 = 0.
Example:
2x² – 3x + 1 = 0 (standard or general quadratic equation )
x² – 9 = 0 (pure quadratic equation)

Methods of Solving Quadratic Equations

Quadratic equations can be solved using different methods, including:

(i) Factorization Method

In this method, we express the quadratic equation as a product of two linear factors.

Example:

Solve x² – 5x + 6 = 0.
Factorizing:
(x – 2)(x – 3) = 0
So, x = 2 or x = 3.

(ii) Completing the Square Method

This involves rewriting the equation in the form of a perfect square trinomial.

Example:

Solve x² + 6x + 5 = 0 by completing the square.
Rewriting:
x² + 6x = -5
Adding (6/2)² = 9 on both sides:
(x + 3)² = 4
Taking square root:
x + 3 = ±2
x = -3 ± 2
So, x = -1 or x = -5.

Quadratic Formula

The quadratic formula is a mathematical formula used to find the roots (solutions) of a quadratic equation of the form: 𝑎𝑥² + 𝑏𝑥 + 𝑐 = 0
The formula is given by: \(x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\)
where:
𝑎, 𝑏, and 𝑐 are the coefficients of the quadratic equation,
± indicates that there are two possible solutions (one with + and one with −),
The expression under the square root, 𝑏² − 4𝑎𝑐, is called the discriminant and determines the nature of the roots.

What is the Quadratic Formula Used For?
The quadratic formula is used to:

1. Solve Quadratic Problems in Physics and Engineering – The formula is widely used in kinematics, projectile motion, circuit analysis, and many other real-world applications where quadratic equations arise.
2. Find the Roots of Any Quadratic Equation – Unlike factorization, which only works when the equation can be factored easily, the quadratic formula works for all quadratic equations, whether they have real or complex roots.
3. Determine the Nature of Roots –
   By evaluating the discriminant (𝑏² − 4𝑎𝑐):
   If 𝑏² − 4𝑎𝑐 > 0 → The equation has two distinct real roots.
   If 𝑏² − 4𝑎𝑐 = 0 → The equation has one real root (repeated).
   If 𝑏² − 4𝑎𝑐 < 0 → The equation has two complex (imaginary) roots.

Example:

Solve 2x² – 4x – 6 = 0 using the quadratic formula.
Here, a = 2, b = -4, c = -6.
\(x = \frac{-(-4) \pm \sqrt{(-4)^2 – 4(2)(-6)}}{2(2)}\)
\(x = \frac{4 \pm \sqrt{16 + 48}}{4}\)
\(x = \frac{4 \pm \sqrt{64}}{4}\)
\(x = \frac{4 \pm 8}{4}\)
x = 3 or x = -1.

Equations Reducible to Quadratic Form

Some equations, though not initially quadratic, can be transformed into quadratic form:

(i) Equations of the type ax⁴ + bx² + c = 0

These are equations in the form of quartic (degree 4) equations, but they can be transformed into quadratic equations by making a substitution.

Substitution: Let 𝑦 = 𝑥².
This substitution reduces the quartic equation to a quadratic equation in 𝑦.

Example:
Solve 2𝑥⁴ − 3𝑥² + 5 = 0.

Let 𝑦 = 𝑥², then the equation becomes:
2𝑦² − 3𝑦 + 5 = 0
Now this equation can easily be solved using quadratic formula.

(ii) Equations of the type ap(x) + b/p(x) = c

Let \( y = p(x) \), then solve the quadratic equation in terms of ( y ).

Example:
Solve \( 2p(x) + \frac{3}{p(x)} = 4 \).

Let \( y = p(x) \), then the equation becomes:
\(2y + \frac{3}{y} = 4.\)
Now, multiply through by ( y ) to eliminate the denominator
\(2y^2 + 3 = 4y.\)
Rearrange the terms to form a quadratic equation:
\(2y^2 – 4y + 3 = 0.\)
Now, solve this quadratic equation for ( y ) using factorization or the quadratic formula.

After finding ( y ), substitute back to find \( p(x) \).

(iii) Reciprocal Equations:

a(x² + 1/x²) + b(x + 1/x) + c = 0
These equations involve terms with both \( x^2 \) and \( \frac{1}{x^2} \), as well as terms with ( x ) and \( \frac{1}{x} \). To solve these equations, we often make a substitution to simplify the equation.

Let \( y = x + \frac{1}{x} \). Then, the equation becomes:

\(a \left( y^2 – 2 \right) + b y + c = 0.\)

Now, expand and simplify the equation:

\(a y^2 – 2a + b y + c = 0.\)

Rearrange the terms:

\(a y^2 + b y + (c – 2a) = 0.\)

Now, solve the quadratic equation for \( y \). Once \( y \) is found, substitute back to find \( x \) using the relation \( y = x + \frac{1}{x} \).

(iv) Exponential Equations

Exponential equations can sometimes be transformed into quadratic form by making an appropriate substitution. For example, if the equation involves terms like \( a^{x} \) or \( b^{x} \), we can set \( y = a^{x} \) or \( y = b^{x} \) to reduce the equation to a quadratic equation in terms of \( y \).

Example:
Solve \( 4^{x} + 2 \cdot 4^{x/2} – 12 = 0 \).

Let \( y = 4^{x/2} \), then \( y^2 = 4^{x} \). Substituting into the equation, we get:

\(y^2 + 2y – 12 = 0.\)

Now, solve this quadratic equation for \( y \), and then back-substitute to find \( x \).

(v) Equations of the type:

\((x + a)(x + b)(x + c)(x + d) = k\)
where \( a + b = c + d \).

This type of equation involves the product of four binomials. By making use of the given condition \( a + b = c + d \), the equation can be reduced to a quadratic form.

Example:
Solve \( (x + 1)(x + 2)(x + 3)(x + 4) = 16 \), where \( 1 + 4 = 2 + 3 \).

First, express the equation as:

\((x + 1)(x + 4) (x + 2)(x + 3)=16\)
Now expand both sides:
\((x^2 + 5x + 4)(x^2 + 5x + 6)=16.\)
let \(y=x^2 + 5x\) then the equation becomes
\((y+4)(y+6)=16\)
further expansion results in \(y^2 +10y+24=16\)
this can be rewritten as \(y^2 + 10y +8=0\)
now this equation in \(y\) can easily be solved using quadratic formula.

Radical Equations

Radical equations involve square roots and can be solved by squaring both sides carefully.

(i) Equations of the type:

\( \sqrt{ax + b} = cx + d \)

These types of equations involve a square root term on one side of the equation. To solve these, first isolate the square root term and then square both sides to remove the square root.
Solve \( \sqrt{2x + 5} = x + 3 \).

Step 1: Isolate the square root term:
\( \sqrt{2x + 5} = x + 3 \).

Step 2: Square both sides to remove the square root:
\( 2x + 5 = (x + 3)^2 \).

Step 3: Expand the right-hand side:
\( 2x + 5 = x^2 + 6x + 9 \).

Step 4: Rearrange the equation to set it equal to zero:
\( x^2 + 6x + 9 – 2x – 5 = 0 \),
\( x^2 + 4x + 4 = 0 \).

Step 5: Factor the quadratic equation:
\( (x + 2)(x + 2) = 0 \).

Step 6: Solve for \( x \):
\( x = -2 \).

(ii) Equations of the type:

\( \sqrt{x + a} + \sqrt{x + b} = \sqrt{x + c} \)

In these types of equations, there are multiple square roots. To solve them, isolate one square root term first, then square both sides to remove the radicals.

Example
Solve \( \sqrt{x + 2} + \sqrt{x + 3} = \sqrt{x + 4} \).

Step 1: Isolate one square root term, say \( \sqrt{x + 4} \):
\( \sqrt{x + 4} = \sqrt{x + 2} + \sqrt{x + 3} \).

Step 2: Square both sides:
\( x + 4 = \left( \sqrt{x + 2} + \sqrt{x + 3} \right)^2 \).

Step 3: Expand the right-hand side using the binomial formula:
\( x + 4 = (x + 2) + 2\sqrt{(x + 2)(x + 3)} + (x + 3) \).

Step 4: Simplify the equation:
\( x + 4 = 2x + 5 + 2\sqrt{(x + 2)(x + 3)} \).

Step 5: Rearrange the terms to isolate the square root:
\( – x – 1 = 2\sqrt{(x + 2)(x + 3)} \).

Step 6: Square both sides again:
\( (x + 1)^2 = 4(x + 2)(x + 3) \).

Step 7: Simplify the equation and solve for \( x \).

(iii) Equations of the type:

\( \sqrt{x^2 + px + m} + \sqrt{x^2 + px + n} = q \)

These are more complex radical equations involving binomial expressions under the square roots. To solve them, isolate one of the square roots, square both sides, and then proceed with solving the resulting quadratic equation.

Example:
Solve \( \sqrt{x^2 + 4x + 3} + \sqrt{x^2 + 4x + 5} = 4 \).

Step 1: Isolate one square root term, say \( \sqrt{x^2 + 4x + 5} \):
\( \sqrt{x^2 + 4x + 5} = 4 – \sqrt{x^2 + 4x + 3} \).

Step 2: Square both sides:
\( x^2 + 4x + 5 = (4 – \sqrt{x^2 + 4x + 3})^2 \).

Step 3: Expand the right-hand side and simplify. This will eventually lead to a quadratic equation which can be solved for \( x \).

Note: Always check for extraneous solutions when squaring both sides, as squaring can sometimes introduce solutions that don’t satisfy the original equation.

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Mastering quadratic equations is essential for building a strong foundation in algebra and tackling more complex math problems in higher classes. By understanding different methods like factorization, the quadratic formula, and equations reducible to quadratic form, you’ll be better prepared to solve real-world problems and ace your exams. Keep practicing with the solved examples provided, and don’t hesitate to explore interactive tools to enhance your learning. We also have solved exercises of the whole unit of quadratic equations for class 10 mathematics.