Quadratic Equation Notes For JEE, NIMCET, CUET, NDA With Formulas, Discriminant, Nature Of Roots, Shortcuts, Solved Examples And Important Tricks For Exams.

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Quadratic Equation – Detailed Notes

1. Definition

A quadratic equation is a polynomial equation of degree 2 in one variable and can be written as:
$ax^2 + bx + c = 0$
where $a, b, c$ are real numbers and $a \neq 0$.

Example: $3x^2 - 5x + 2 = 0$

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2. Standard Form and Types

2.1 Standard (General) Form

$ax^2 + bx + c = 0$

2.2 Pure Quadratic Equation

When $b = 0$: $ax^2 + c = 0$

Example: $5x^2 - 20 = 0$

2.3 Incomplete Quadratic Equation

• Without constant term: $ax^2 + bx = 0 \Rightarrow x(ax + b) = 0$
• Without $x$ term: $ax^2 + c = 0$

2.4 Quadratic in Disguise

Sometimes equation can be reduced to quadratic form by substitution.

Example: $t^4 - 5t^2 + 4 = 0$
Let $y = t^2$ then $y^2 - 5y + 4 = 0$ (quadratic in $y$).

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3. Solutions (Roots) of a Quadratic Equation

The values of $x$ which satisfy $ax^2 + bx + c = 0$ are called the roots or solutions.

3.1 Factorisation Method

Steps:

1. Multiply $a \times c$.
2. Split $b$ into two parts whose product is $ac$ and sum is $b$.
3. Group terms and factorise.
4. Equate each factor to zero.

Example: Solve $x^2 - 7x + 12 = 0$.

$x^2 - 7x + 12 = 0$
Split $-7$ as $-3 - 4$:
$x^2 - 3x - 4x + 12 = 0$
$x(x - 3) - 4(x - 3) = 0$
$(x - 3)(x - 4) = 0$
$\Rightarrow x = 3, 4$

3.2 Method of Completing the Square

Idea: Convert quadratic expression into a perfect square.

Example: Solve $x^2 + 6x + 5 = 0$.

$x^2 + 6x + 5 = 0$
$x^2 + 6x = -5$
Add $\left(\frac{6}{2}\right)^2 = 9$ on both sides:
$x^2 + 6x + 9 = 4$
$(x + 3)^2 = 4$
$x + 3 = \pm 2$
$x = -3 + 2 = -1$ or $x = -3 - 2 = -5$

3.3 Quadratic Formula (General Formula)

For $ax^2 + bx + c = 0, a \neq 0$ the roots are given by:
$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Example: Solve $2x^2 + 3x - 2 = 0$.

Here $a = 2, b = 3, c = -2$
$D = b^2 - 4ac = 3^2 - 4(2)(-2) = 9 + 16 = 25$
$x = \dfrac{-3 \pm \sqrt{25}}{2 \times 2} = \dfrac{-3 \pm 5}{4}$
$x_1 = \dfrac{-3 + 5}{4} = \dfrac{2}{4} = \dfrac{1}{2}$
$x_2 = \dfrac{-3 - 5}{4} = \dfrac{-8}{4} = -2$

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4. Discriminant and Nature of Roots

For $ax^2 + bx + c = 0$, the discriminant is:
$D = b^2 - 4ac$

Value of $D$	Nature of Roots
$D > 0$ and perfect square	Two real, distinct and rational roots
$D > 0$ but not a perfect square	Two real, distinct and irrational roots
$D = 0$	Two real and equal roots (repeated root)
$D < 0$	No real roots, two complex conjugate roots

Example: $x^2 - 4x + 4 = 0$

$a = 1, b = -4, c = 4$
$D = (-4)^2 - 4(1)(4) = 16 - 16 = 0$
Roots are real and equal.

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5. Relationship Between Roots and Coefficients

Let $ax^2 + bx + c = 0$ have roots $\alpha$ and $\beta$. Then:

• Sum of roots: $\alpha + \beta = -\dfrac{b}{a}$
• Product of roots: $\alpha \beta = \dfrac{c}{a}$

5.1 Forming a Quadratic Equation from Given Roots

If roots are $\alpha$ and $\beta$, then equation is:

$x^2 - (\alpha + \beta)x + \alpha \beta = 0$

Example: Roots are $3$ and $5$.

$\alpha + \beta = 8, \ \alpha \beta = 15$
Equation: $x^2 - 8x + 15 = 0$

5.2 Condition for Equal Roots

For equal (repeated) roots, $D = 0$ i.e. $b^2 - 4ac = 0$.

5.3 Condition for Real Roots

For real roots (distinct or equal), $D \geq 0$.

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6. Special Cases

6.1 When $c = 0$

Equation: $ax^2 + bx = 0$

$x(ax + b) = 0 \Rightarrow x = 0$ or $x = -\dfrac{b}{a}$

6.2 When $b = 0$

Equation: $ax^2 + c = 0$

$ax^2 = -c \Rightarrow x^2 = -\dfrac{c}{a}$
If $-\dfrac{c}{a} > 0$ roots are real; if $-\dfrac{c}{a} < 0$ roots are imaginary.

6.3 Quadratic in $x^2$

Equations like $x^4 + 5x^2 + 4 = 0$ can be solved by substitution $y = x^2$.

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7. Graph of a Quadratic Equation

The graph of $y = ax^2 + bx + c$ is a parabola.

• If $a > 0$ parabola opens upwards.
• If $a < 0$ parabola opens downwards.

7.1 Vertex

Vertex of parabola $y = ax^2 + bx + c$ is given by:
$x_v = -\dfrac{b}{2a}$, and $y_v = f(x_v)$

7.2 Axis of Symmetry

Line $x = -\dfrac{b}{2a}$ is the axis of symmetry of the parabola.

7.3 Relation Between Graph and Nature of Roots

• Parabola cuts x-axis at two distinct points $\Rightarrow$ two distinct real roots.
• Parabola touches x-axis at one point (tangent) $\Rightarrow$ one real repeated root.
• Parabola does not cut x-axis $\Rightarrow$ no real roots (only complex roots).

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8. Maximum and Minimum Value of a Quadratic Expression

For $y = ax^2 + bx + c$:

• If $a > 0$, $y$ has a minimum value at $x = -\dfrac{b}{2a}$.
• If $a < 0$, $y$ has a maximum value at $x = -\dfrac{b}{2a}$.

The extremum (max or min) value is given by:
$y_{\min/\max} = c - \dfrac{b^2}{4a}$

Example: Find minimum value of $y = x^2 - 4x + 7$.

Here $a = 1 > 0$ so minimum exists.
$x_v = -\dfrac{b}{2a} = -\dfrac{-4}{2} = 2$
$y_{\min} = 1(2)^2 - 4(2) + 7 = 4 - 8 + 7 = 3$

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9. Applications of Quadratic Equations

9.1 Word Problems

• Motion under gravity (projectile, height-time relation).
• Area problems (length and breadth with conditions).
• Number problems (consecutive integers, age problems).
• Profit–revenue–cost models (maximizing profit).

Example (Number Problem):

The product of two consecutive natural numbers is $156$. Find the numbers.
Let numbers be $n$ and $n+1$.
$n(n+1) = 156 \Rightarrow n^2 + n - 156 = 0$
Solve using formula or factorisation to get $n = 12$ or $n = -13$ (reject negative).
So numbers are $12$ and $13$.

Example (Area Problem):

A rectangle has length $(x+3)$ and breadth $(x-2)$. If area is $80$, find $x$.
$(x+3)(x-2) = 80$
$x^2 + x - 6 = 80$
$x^2 + x - 86 = 0$
Solve using quadratic formula.

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10. Important Results and Quick Points

• General form: $ax^2 + bx + c = 0, a \neq 0$.
• Discriminant: $D = b^2 - 4ac$.
• Roots: $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Sum of roots: $\alpha + \beta = -\dfrac{b}{a}$.
• Product of roots: $\alpha \beta = \dfrac{c}{a}$.
• $D > 0$ $\Rightarrow$ two distinct real roots.
• $D = 0$ $\Rightarrow$ real and equal roots.
• $D < 0$ $\Rightarrow$ no real roots (complex roots).
• For equal roots: $b^2 = 4ac$.
• Parabola opens up if $a > 0$, down if $a < 0$.

11. Location of Roots of a Quadratic Equation

Consider the quadratic equation $ax^2 + bx + c = 0$ with real roots $\alpha$ and $\beta$. We know:

• Sum of roots: $\alpha + \beta = -\dfrac{b}{a}$
• Product of roots: $\alpha \beta = \dfrac{c}{a}$

For any real number $k$, define $f(x) = ax^2 + bx + c$. Then: $$(\alpha - k)(\beta - k) = \dfrac{f(k)}{a} = \dfrac{ak^2 + bk + c}{a}$$ This helps to decide where the roots lie on the number line.

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11.1 Location of Roots with Respect to Zero (Sign of Roots)

(A) Both Roots Positive

Both roots $\alpha, \beta > 0$ if:

• $D = b^2 - 4ac \ge 0$ (roots are real),
• $\alpha + \beta = -\dfrac{b}{a} > 0$ (sum positive),
• $\alpha \beta = \dfrac{c}{a} > 0$ (product positive).

(B) Both Roots Negative

Both roots $\alpha, \beta < 0$ if:

• $D \ge 0$,
• $\alpha + \beta = -\dfrac{b}{a} < 0$ (sum negative),
• $\alpha \beta = \dfrac{c}{a} > 0$ (product positive).

(C) Roots of Opposite Signs

One root positive and one root negative if:

• $D > 0$ (two distinct real roots),
• $\alpha \beta = \dfrac{c}{a} < 0$ (product negative).

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11.2 Location of Roots with Respect to a Point $x = k$

(i) Roots on Opposite Sides of $k$

One root less than $k$ and the other greater than $k$ (i.e. $\alpha < k < \beta$ or $\beta < k < \alpha$) if:

• $D > 0$,
• $(\alpha - k)(\beta - k) < 0 \;\Longleftrightarrow\; \dfrac{f(k)}{a} < 0 \;\Longleftrightarrow\; f(k)\cdot a < 0$.

In coefficient form: roots lie on opposite sides of $k$ if $$f(k) = ak^2 + bk + c \text{ and } f(k)\cdot a < 0.$$

(ii) Both Roots Greater Than $k$

Both $\alpha, \beta > k$ if:

• $D \ge 0$ (real roots),
• $(\alpha - k)(\beta - k) > 0 \;\Longleftrightarrow\; \dfrac{f(k)}{a} > 0$ (both to the same side of $k$),
• $\alpha + \beta > 2k \;\Longleftrightarrow\; -\dfrac{b}{a} > 2k$ (their average lies to the right of $k$).

So the conditions are: $$D \ge 0,\quad \dfrac{f(k)}{a} > 0,\quad -\dfrac{b}{a} > 2k.$$

(iii) Both Roots Less Than $k$

Both $\alpha, \beta < k$ if:

• $D \ge 0$,
• $(\alpha - k)(\beta - k) > 0 \;\Longleftrightarrow\; \dfrac{f(k)}{a} > 0$,
• $\alpha + \beta < 2k \;\Longleftrightarrow\; -\dfrac{b}{a} < 2k$.

So the conditions are: $$D \ge 0,\quad \dfrac{f(k)}{a} > 0,\quad -\dfrac{b}{a} < 2k.$$

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11.3 Special Case: Roots Around a Point $k$

Exactly One Root Equal to $k$

• $k$ is a root $\Longleftrightarrow f(k) = 0$.

Both Roots on One Side of $k$

• Both roots $\ge k$ if $f(k) \ge 0$ and $f(x)$ does not change sign for $x \ge k$ (check vertex position).
• Both roots $\le k$ if $f(k) \ge 0$ and $f(x)$ does not change sign for $x \le k$.

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11.4 Location of Roots Inside an Interval $(p, q)$

Let $p < q$. For both roots $\alpha, \beta$ to lie in $(p, q)$:

• $D > 0$ (two distinct real roots),
• $f(p)$ and $f(q)$ have the same sign and are opposite in sign to $a$ (parabola cuts x–axis between $p$ and $q$),
• Vertex $x_v = -\dfrac{b}{2a}$ lies between $p$ and $q$.

In summary, for both roots in $(p, q)$: $$p < -\dfrac{b}{2a} < q,\quad f(p)\cdot a > 0,\quad f(q)\cdot a > 0.$$

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11.5 Quick Summary (For Problems)

• Roots of opposite sign $\Longleftrightarrow \dfrac{c}{a} < 0$.
• Both roots positive $\Longleftrightarrow D \ge 0,\ \dfrac{c}{a} > 0,\ -\dfrac{b}{a} > 0$.
• Both roots negative $\Longleftrightarrow D \ge 0,\ \dfrac{c}{a} > 0,\ -\dfrac{b}{a} < 0$.
• Roots on opposite sides of $k$ $\Longleftrightarrow f(k)\cdot a < 0$.
• Both roots > $k$ $\Longleftrightarrow D \ge 0,\ \dfrac{f(k)}{a} > 0,\ -\dfrac{b}{a} > 2k$.
• Both roots < $k$ $\Longleftrightarrow D \ge 0,\ \dfrac{f(k)}{a} > 0,\ -\dfrac{b}{a} < 2k$.

Cubic Equation – Detailed Notes

1. Definition

A cubic equation is a polynomial equation of degree 3 and is written as:
$ax^3 + bx^2 + cx + d = 0$
where $a, b, c, d$ are real numbers and $a \neq 0$.

Example: $2x^3 - 3x^2 - 11x + 6 = 0$

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2. Types of Cubic Equation

2.1 General Form

$ax^3 + bx^2 + cx + d = 0$

2.2 Depressed Cubic

A cubic without the $x^2$ term is called a depressed cubic. It can be obtained using substitution $x = y - \dfrac{b}{3a}$.

Form: $y^3 + py + q = 0$

2.3 Special Cubic

• Factorable cubic — can be broken into linear/quadratic factors.
• Perfect cube type: $(x - a)^3 = 0$
• Symmetric cubic: $ax^3 + bx^2 - bx - a = 0$

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3. Methods of Solving Cubic Equations

3.1 Factorisation Method

If one obvious root is found, divide the cubic by $(x - \alpha)$.

Example:

Solve $x^3 - 6x^2 + 11x - 6 = 0$ Try small values: $x = 1$ is a root.
Divide by $(x - 1)$:
$x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6)$
Factorise quadratic:
$x^2 - 5x + 6 = (x - 2)(x - 3)$
Roots: $x = 1, 2, 3$

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3.2 Using Rational Root Theorem

Possible rational roots are factors of constant term $d$ divided by factors of leading term $a$.

Example: For $2x^3 - 3x^2 - 11x + 6 = 0$

Possible roots = $\pm 1, \pm 2, \pm 3, \pm 6, \pm \dfrac{1}{2}, \pm \dfrac{3}{2}$
Testing gives $x = 3$ as a root.

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3.3 Synthetic Division / Long Division

Once one root is known, divide the cubic to reduce it to quadratic.

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3.4 Cardano’s Method (Formula Method)

For the depressed cubic: $y^3 + py + q = 0$ Roots are given by:

$y = \sqrt[3]{-\dfrac{q}{2} + \sqrt{\Delta}} + \sqrt[3]{-\dfrac{q}{2} - \sqrt{\Delta}}$ where $\Delta = \left(\dfrac{q}{2}\right)^2 + \left(\dfrac{p}{3}\right)^3$

Not used at school level due to complexity.

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4. Nature of Roots

The discriminant of a cubic is:

$\Delta = 18abcd - 4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2$

• Δ > 0 → Three distinct real roots
• Δ = 0 → Multiple roots; at least two equal
• Δ < 0 → One real and two complex conjugate roots

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5. Relation Between Roots and Coefficients

Let roots be $\alpha, \beta, \gamma$ of $ax^3 + bx^2 + cx + d = 0$.

• Sum of roots: $\alpha + \beta + \gamma = -\dfrac{b}{a}$
• Sum of product of roots taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = \dfrac{c}{a}$
• Product of roots: $\alpha\beta\gamma = -\dfrac{d}{a}$

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6. Finding a Cubic Equation from Given Roots

If roots are $\alpha, \beta, \gamma$ then the equation is:

$x^3 - (\alpha + \beta + \gamma)x^2 + (\alpha\beta + \beta\gamma + \gamma\alpha)x - \alpha\beta\gamma = 0$

Example:

Roots are $1, 2, 3$
Sum = 6 Sum of pairwise products = $1\cdot2 + 2\cdot3 + 3\cdot1 = 11$ Product = 6 Equation: $x^3 - 6x^2 + 11x - 6 = 0$

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7. Graph of a Cubic Function

The graph of $y = ax^3 + bx^2 + cx + d$ is called a cubic curve.

Key Features:

• Always has at least one real root (curve must cross x-axis).
• If $a > 0$, the right end rises and the left end falls.
• If $a < 0$, the right end falls and the left end rises.
• Can have 1 or 3 real roots.
• May have turning points (maxima/minima).

Turning Points:

Given by solving $y' = 0$ for $y = ax^3 + bx^2 + cx + d$.
$y' = 3ax^2 + 2bx + c$

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8. Applications of Cubic Equations

• Volume and geometry problems
• Motion under variable acceleration
• Profit–cost modelling
• Physics: equilibrium problems
• Population and economic modelling

Example (Volume Problem):

A box has a volume of 120 cm³. If its length is $(x+1)$, breadth $(x-2)$ and height $x$, then: $x(x+1)(x-2) = 120$ which leads to a cubic equation.

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9. Important Points to Remember

• Cubic equation always has at least one real root.
• Try Rational Root Theorem first.
• If one root is known → reduce to quadratic.
• Use formulas for sum/product of roots for quick checks.
• Graph always crosses the x-axis at least once.
• Discriminant helps predict number of real roots.

Rolle's Theorem – Detailed Notes with Examples

1. Definition

Rolle's Theorem is a fundamental result in calculus. It states that for a function $f(x)$ defined on the closed interval $[a, b]$, if:

1. $f(x)$ is continuous on the closed interval $[a, b]$
2. $f(x)$ is differentiable on the open interval $(a, b)$
3. $f(a) = f(b)$

Then there exists at least one $c$ in $(a, b)$ such that: $$f'(c) = 0$$

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2. Geometrical Meaning

If a smooth curve starts and ends at the same height at $x=a$ and $x=b$, then there must be at least one point between them where the tangent is horizontal (slope = 0).

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3. Conditions of Rolle’s Theorem

• Continuity on $[a, b]$ ensures no breaks or jumps.
• Differentiability on $(a, b)$ ensures smoothness (no corners/cusps).
• Equal values at endpoints ensures the curve returns to the same level.

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4. Example Problems (Fully Solved)

Example 1

Verify Rolle's Theorem for $f(x) = x^2 - 4x + 3$ on $[1, 3]$.

Step 1: Check continuity and differentiability
$f(x)$ is a polynomial → continuous and differentiable everywhere.

Step 2: Check $f(a) = f(b)$
$f(1) = 1 - 4 + 3 = 0$
$f(3) = 9 - 12 + 3 = 0$
So $f(1) = f(3)$.

Step 3: Find $c$ such that $f'(c) = 0$
$f'(x) = 2x - 4$ Set $f'(c) = 0$:
$2c - 4 = 0$ → $c = 2$

Answer: Rolle’s theorem verified, $c = 2$.

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Example 2

Verify Rolle's Theorem for $f(x) = \sin x$ on $[0, \pi]$.

Step 1: $\sin x$ is continuous and differentiable everywhere.

Step 2: $f(0) = 0$, $f(\pi) = 0$ → equal.

Step 3:
$f'(x) = \cos x$
Set $\cos c = 0$ → $c = \frac{\pi}{2}$

Answer: $c = \dfrac{\pi}{2}$.

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Example 3

Verify Rolle's Theorem for $f(x) = x^3 - 3x$ on $[-\sqrt{3}, \sqrt{3}]$.

Step 1: Polynomial → continuous and differentiable.

Step 2:
$f(-\sqrt{3}) = -3\sqrt{3} + 3\sqrt{3} = 0$
$f(\sqrt{3}) = 3\sqrt{3} - 3\sqrt{3} = 0$

Step 3:
$f'(x) = 3x^2 - 3$
$3c^2 - 3 = 0$ → $c^2 = 1$ → $c = \pm 1$

Answer: $c = -1, 1$.

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Example 4

Verify Rolle's Theorem for $f(x) = x \ln x$ on $[1, e]$.

Step 1: Function continuous on $[1, e]$, differentiable on $(1, e)$.

Step 2:
$f(1) = 1 \ln 1 = 0$
$f(e) = e \ln e = e(1) = e$ Not equal → Rolle's Theorem NOT applicable.

Conclusion: Rolle's theorem fails since $f(1) \neq f(e)$.

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Example 5

Verify Rolle's Theorem for $f(x) = |x|$ on $[-1, 1]$.

Step 1: $|x|$ is continuous everywhere → OK.

Step 2: $f(-1) = 1$, $f(1) = 1$ → equal.

Step 3: Check differentiability:
$f(x)$ is not differentiable at $x = 0$

Conclusion: Rolle's theorem does NOT apply because $|x|$ is not differentiable on $(a, b)$.

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5. Summary

• Rolle's theorem guarantees at least one $c$ in $(a, b)$ where $f'(c) = 0$.
• Applicable only if all three conditions are satisfied.
• Useful in proving the Mean Value Theorem.
• Applicable mostly to smooth functions like polynomials, trigonometric and exponential functions.

Common Roots of Algebraic Equations – Detailed Notes

1. Definition

Two algebraic equations are said to have a common root if there exists at least one value of $x$ that satisfies both equations simultaneously.

Example: If $x = 2$ satisfies both equations, then 2 is a common root.

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2. General Approach to Find Common Root(s)

Method 1: Elimination (Subtract the equations)

• Subtract one equation from another to eliminate $x^2$, $x$, etc.
• The result is usually a linear equation → gives common root.

Method 2: Using Resultant

If two quadratic equations share a common root, their resultant = 0. But this method is rarely used at school level.

Method 3: Express $x$ from one equation and substitute in the other

Useful for linear–quadratic combinations.

Method 4: Using Sum and Product of Roots

If $\alpha$ is common root of two quadratics:
$ax^2 + bx + c = 0$ and $px^2 + qx + r = 0$, then plugging $\alpha$ in both gives: $$a\alpha^2 + b\alpha + c = 0$$ $$p\alpha^2 + q\alpha + r = 0$$ Subtract to eliminate $\alpha^2$: $$(b - q)\alpha + (c - r) = 0$$ Solve for $\alpha$.

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3. Condition for Common Roots (Important Formula)

For two quadratics $ax^2 + bx + c = 0$ $px^2 + qx + r = 0$ to have a common root, the determinant must vanish:

$$ \begin{vmatrix} a & b & c \\ p & q & r \\ a+p & b+q & c+r \end{vmatrix} = 0 $$

But for exams like JEE/NIMCET, simpler elimination is preferred.

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4. Examples

Example 1

Find the common root of
$x^2 - 5x + 6 = 0$ and $x^2 - 7x + 12 = 0$.

Step 1: Subtract equations
$(x^2 - 5x + 6) - (x^2 - 7x + 12) = 0$
$2x - 6 = 0$

Step 2: Solve
$2x = 6$ → $x = 3$

Answer: Common root = 3 Check: In first eq → $3^2 - 15 + 6 = 0$ In second eq → $9 - 21 + 12 = 0$

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Example 2

Find common root of
$2x^2 - 9x + 4 = 0$ and $3x^2 - 14x + 8 = 0$.

Subtract:
$(2x^2 - 9x + 4) - (3x^2 - 14x + 8) = 0$
$-x^2 + 5x - 4 = 0$

Factor:
$-(x^2 - 5x + 4) = 0$ $x^2 - 5x + 4 = 0$ $x = 1, 4$

Check which one is common:

$x = 1$ does NOT satisfy second equation. $x = 4$ gives: $2(16) - 36 + 4 = 0$ $3(16) - 56 + 8 = 0$ So common root = 4

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Example 3

Find common root of:
$x^2 - 4 = 0$ and $x^2 - 9x + 20 = 0$.

First equation: $x^2 - 4 = 0$ → $x = 2, -2$

Check in second equation:

For $x = 2$: $4 - 18 + 20 = 6 \neq 0$ For $x = -2$: $4 + 18 + 20 = 42 \neq 0$

Conclusion: No common root.

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Example 4

Find common root of
$x^2 + 3x - 10 = 0$ and $2x^2 + x - 15 = 0$.

Subtract:
$(x^2 + 3x - 10) - (2x^2 + x - 15)$
$-x^2 + 2x + 5 = 0$ → $x^2 - 2x - 5 = 0$

Quadratic formula:

$x = \dfrac{2 \pm \sqrt{4 + 20}}{2}$ $x = 1 \pm \sqrt{6}$

Check which is common root (substitute back): Only $1 + \sqrt{6}$ satisfies both equations.

Answer: Common root = $1 + \sqrt{6}$

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Example 5

Find common root of
$x^2 - 3x + 2 = 0$ and $x^2 - 5x + 6 = 0$.

Factor first: $(x-1)(x-2)=0$ → roots: 1 and 2

Factor second: $(x-2)(x-3)=0$ → roots: 2 and 3

Common root: 2

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5. Summary

• Common root means root satisfying both equations.
• Best method: subtract one equation from the other.
• If subtraction gives linear → easy to find common root.
• Always recheck by substituting in both equations.
• Two quadratics can have 0, 1 or 2 common roots.

Transformation of Equations – Detailed Notes

1. Meaning of Transformation

Transformation of an equation means converting an existing algebraic equation into another equation whose roots are related to the roots of the original equation in a specific way (such as increased, decreased, multiplied, divided, reciprocals, squared, etc.).

If an original equation is:
$ax^n + bx^{n-1} + cx^{n-2} + \cdots + k = 0$ and its roots are $\alpha_1, \alpha_2, \ldots, \alpha_n$, then after transformation we obtain a new equation whose roots are some function of these roots.

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2. Standard Transformations

2.1 When the Roots are Increased by a Constant $h$

If new roots = $\alpha + h$, then substitute:

$x = y - h$

Replace $x$ in the original equation with $(y - h)$ and simplify to get the new equation in $y$.

Example:
Transform $x^2 - 5x + 6 = 0$ so roots become $(\alpha + 2)$.
Use $x = y - 2$:

$(y - 2)^2 - 5(y - 2) + 6 = 0$
$y^2 - 4y + 4 - 5y + 10 + 6 = 0$
$y^2 - 9y + 20 = 0$

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2.2 When the Roots are Decreased by a Constant $h$

New roots = $\alpha - h$ Use substitution:

$x = y + h$

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2.3 When the Roots are Multiplied by a Constant $k$

New roots = $k\alpha$ Use substitution:

$x = \frac{y}{k}$ Multiply by $k^n$ to clear denominator.

Example:
Roots multiplied by 3 for $x^2 - 4x + 1 = 0$:

Use $x = y/3$:
$(y/3)^2 - 4(y/3) + 1 = 0$
Multiply by 9:
$y^2 - 12y + 9 = 0$

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2.4 When the Roots are Divided by $k$

New roots = $\dfrac{\alpha}{k}$ Substitution:

$x = ky$

Expand and simplify.

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2.5 When the New Roots are Reciprocals ($\dfrac{1}{\alpha}$)

Take original equation: $ax^n + bx^{n-1} + \cdots + k = 0$

Substitute $x = \frac{1}{y}$:

$a(1/y)^n + b(1/y)^{n-1} + \cdots + k = 0$

Multiply by $y^n$ to get new equation in $y$.

Shortcut:
Reverse the coefficients if leading coefficient = constant term (for monic symmetric polynomials).

Example:
Find equation whose roots are reciprocals of $x^2 - 5x + 6 = 0$:

$6x^2 - 5x + 1 = 0$ (reverse coefficients: 6, -5, 1)

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2.6 When New Roots are Squares ($\alpha^2$)

If original roots = $\alpha, \beta$ then new quadratic has roots $\alpha^2$ and $\beta^2$.

Formula:

• Sum of new roots = $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
• Product of new roots = $\alpha^2 \beta^2 = (\alpha\beta)^2$

Example:
Original: $x^2 - 5x + 6 = 0$ Sum = 5, Product = 6 New roots: $\alpha^2, \beta^2$

Sum = $5^2 - 2(6) = 25 - 12 = 13$
Product = $6^2 = 36$

New equation: $x^2 - 13x + 36 = 0$

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2.7 Roots Equal to Square Root of Original Roots ($\sqrt{\alpha}$)

Works only when original roots are positive. If roots are $\alpha, \beta$ then new equation has roots $\sqrt{\alpha}, -\sqrt{\alpha}, \sqrt{\beta}, -\sqrt{\beta}.$

Degree becomes doubled.

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2.8 Roots Equal to Negative of Original Roots ($-\alpha$)

Substitution: $x = -y$

Example:
Transform $x^2 - 4x + 3 = 0$ so roots become $-\alpha$:

$x = -y$:
$(-y)^2 - 4(-y) + 3 = 0$
$y^2 + 4y + 3 = 0$

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3. Important Shortcut Table

New Roots	Substitution	Method
$\alpha + h$	$x = y - h$	Expand
$\alpha - h$	$x = y + h$	Expand
$k\alpha$	$x = y/k$	Multiply by $k^n$
$\alpha/k$	$x = ky$	Expand
$1/\alpha$	$x = 1/y$	Reverse coefficients
$\alpha^2$	Use sum/product formulas	New quadratic

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4. Practice Examples (Solved)

Example 1

Transform $x^2 - 7x + 10 = 0$ so that roots become $(\alpha + 3)$.

$x = y - 3$:
$(y - 3)^2 - 7(y - 3) + 10 = 0$
$y^2 - 6y + 9 - 7y + 21 + 10 = 0$
$y^2 - 13y + 40 = 0$

Example 2

Find equation whose roots are triple the roots of $x^2 - 4x + 1 = 0$.

$x = y/3$:
$(y/3)^2 - 4(y/3) + 1 = 0$
Multiply by 9:
$y^2 - 12y + 9 = 0$

Example 3

Find equation whose roots are reciprocals of $2x^2 - 5x + 3 = 0$.

Reverse coefficients:
$3x^2 - 5x + 2 = 0$

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5. Summary

• Transformations modify the roots by substitution.
• Most common transformations: increase, decrease, multiply, divide, reciprocals.
• Use substitution method for all except reciprocals (use reverse method).
• Use relation of roots for squared root transformations.