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❄️Notes on Quadratic Equations⛄️


In order to solve a quadratic equation of the form ax2 + bx + c, we first need to calculate the discriminant with the help of the formula D = b2 – 4ac.

The solution of the quadratic equation ax2 + bx + c= 0 is given by x = [-b ± √ b2 – 4ac] / 2a

If α and β are the roots of the quadratic equation ax2 + bx + c = 0, then we have the following results for the sum and product of roots:

α + β = -b/a

α.β = c/a

α – β = √D/a

It is not possible for a quadratic equation to have three different roots and if in any case it happens, then the equation becomes an identity.

Nature of Roots:

Consider an equation ax2 + bx + c = 0, where a, b and c ∈ R and a ≠ 0, then we have the following cases:

D > 0 iff the roots are real and distinct i.e. the roots are unequal

D = 0 iff the roots are real and coincident i.e. equal

D < 0 iffthe roots are imaginary

The imaginary roots always occur in pairs i.e. if a+ib is one root of a quadratic equation, then the other root must be the conjugate i.e. a-ib, where a, b ∈ R and i = √-1.

Consider an equation ax2 + bx + c = 0, where a, b and c ∈Q and a ≠ 0, then

If D > 0 and is also a perfect square then the roots are rational and unequal.

If α = p + √q is a root of the equation, where ‘p’ is rational and √q is a surd, then the other root must be the conjugate of it i.e. β = p - √q and vice versa.

If the roots of the quadratic equation are known, then the quadratic equation may be constructed with the help of the formula
x2 – (Sum of roots)x + (Product of roots) = 0.

So if α and β are the roots of equation then the quadratic equation is

x2 – (α + β)x + α β = 0

For the quadratic expressiony = ax2 + bx + c, where a, b, c ∈ R and a ≠ 0, then the graph between x and y is always a parabola.

If a > 0, then the shape of the parabola is concave upwards

If a < 0, then the shape of the parabola is concave upwards

Inequalities of the form P(x)/ Q(x) > 0 can be easily solved by the method of intervals of number line rule.

The maximum and minimum values of the expression y = ax2 + bx + c occur at the point x = -b/2a depending on whether a > 0 or a< 0.

y ∈[(4ac-b2) / 4a, ∞] if a > 0

If a < 0, then y ∈ [-∞, (4ac-b2) / 4a]

The quadratic function of the form f(x, y) = ax2+by2 + 2hxy + 2gx + 2fy + c = 0 can be resolved into two linear factors provided it satisfies the following condition: abc + 2fgh –af2 – bg2 – ch2 = 0

In general, if α1,α2, α3, …… ,αn are the roots of the equation

f(x) = a0xn +a1xn-1 + a2xn-2 + ……. + an-1x + an, then

1.Σα1 = - a1/a0

2.Σ α1α2 = a2/a0

3.Σ α1α2α3 = - a3/a0

……… ……….

Σ α1α2α3 ……αn= (-1)n an/a0

Every equation of nth degree has exactly n roots (n ≥1) and if it has more than n roots then the equation becomes an identity.

If there are two real numbers ‘a’ and ‘b’ such that f(a) and f(b) are of opposite signs, then f(x) = 0 must have at least one real root between ‘a’ and ‘b’.

Every equation f(x) = 0 of odd degree has at least one real root of a sign opposite to that of its last term.
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✍️Revision Notes on Vectors
➖➖➖➖➖➖➖➖➖➖➖➖

The length or the magnitude of the vector = (a, b, c) is defined by w = √a2+b2+c2

A vector may be divided by its own length to convert it into a unit vector, i.e. ? = u / |u|. (The vectors have been denoted by bold letters.)

If the coordinates of point A are xA, yA, zA and those of point B are xB, yB, zB then the vector connecting point A to point B is given by the vector r, where r = (xB - xA)i + (yB – yA) j + (zB – zA)k , here i, j and k denote the unit vectors along x, y and z axis respectively.

Some key points of vectors:
1) The magnitude of a vector is a scalar quantity

2) Vectors can be multiplied by a scalar. The result is another vector.

3) Suppose c is a scalar and v = (a, b) is a vector, then the scalar multiplication is defined by cv = c (a, b) = (ca, cb). Hence each component of vector is multiplied by the scalar.

4) If two vectors are of the same dimension then they can be added or subtracted from each other. The result is gain a vector.

If u, v and w are three vectors and c, d are scalars then the following results of vector addition hold true:
1) u + v = v + u (the commutative law of addition)

2) u + 0 = u

3) u + (-u) = 0 (existence of additive inverses)

4) c (du) = (cd)u

5) (c + d)u = cu + d u

6) c(u + v) = cu + cv

7) 1u = u

8) u + (v + w) = (u + v) + w (the associative law of addition)

Some Basic Rules of Algebra of Vectors:
1) a.a = |a|2 = a2

2) a.b = b.a

3) a.0 = 0

4) a.b = (a cos q)b = (projection of a on b)b = (projection of b on a) a

5) a.(b + c) = a.b + a.c (This is also termed as the distributive law)

6) (la).(mb) = lm (a.b)

7) (a ± b)2 = (a ± b) . (a ± b) = a2 + b2 ± 2a.b

8) If a and b are non-zero, then the angle between them is given by cos θ = a.b/|a||b|

9) a x a = 0

10) a x b = - (b x a)

11) a x (b + c) = a x b + a x c

Any vector perpendicular to the plane of a and b is l(a x b) where l is a real number.

Unit vector perpendicular to a and b is ± (a x b)/ |a x b|

The position of dot and cross can be interchanged without altering the product. Hence it is also represented by [a b c]

1) [a b c] = [b c a] = [c a b]

2) [a b c] = - [b a c]

3) [ka b c] = k[a b c]

4) [a+b c d] = [a c d] + [b c d]

5) a x (b x c) = (a x b) x c, if some or all of a, b and c are zero vectors or a and c are collinear.

Methods to prove collinearity of vectors:
1) Two vectors a and b are said to be collinear if there exists k ? R such that a = kb.

2) If p x q = 0, then p and q are collinear.

3) Three points A(a), B(b) and C(c) are collinear if there exists k ? R such that AB = kBC i.e. b-a = k (c-b).

4) If (b-a) x (c-b) = 0, then A, B and C are collinear.

5) A(a), B(b) and C(c) are collinear if there exists scalars l, m and n (not all zero) such that la + mb+ nc = 0, where l + m + n = o

Three vectors p, q and r are coplanar if there exists l, m ? R such that r = lp + mq i.e., one can be expressed as a linear combination of the other two.

If [p q r] = 0, then p, q and r are coplanar.

Four points A(a), B(b), C(c) and D(d) lie in the same plane if there exist l, m ? R such that b-a = l(c-b) + m(d-c).

If [b-a c-b d-c] = 0 then A, B, C, D are coplanar.

Two lines in space can be parallel, intersecting or neither (called skew lines). Let r = a1 + μb1 and r = a2 + μb2 be two lines.

They intersect if (b1 x b2)(a2 - a1) = 0

The two lines are parallel if b1 and b2 are collinear.

The angle between two planes is the angle between their normal unit vectors i.e. cos q = n1 . n2

If a, b and c are three coplanar vectors, then the system of vectors a', b' and c' is said to be the reciprocal system of vectors if aa' = bb' = cc' = 1 where a' = (b xc) /[a b c] , b' = (c xa)/ [a b c] and c' = (a x b)/[a b c] Also, [a' b' c'] = 1/ [a b c]

Dot Product of two vectors a and b defined by a = [a1, a2, ..., an] and b = [b1, b2, ..., bn] is given by a1b1 + a2b2 + ..., + anbn .
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♦️Trigonometry- Solution of Triangles♦️

Sine rule: Sides of a triangle are proportional to the sine of the angles opposite to them. So, in ΔABC,
sin A/a = sin B/b = sin C/c = 2Δ/abc.

This may also be written as (a/sin A) = (b/sin B) = (c/sin C)

Cosine rule: In any ΔABC,
cos A = (b2 + c2 – a2) /2bc


cos B = (a2 + c2 – b2)/2ac

cos C = (a2 + b2 - c2)/2ab

Trigonometric ratios of half-angles:
sin A/2 = √[(s-b)(s-c)/bc]

sin B/2 = √[(s-c) (s-a)/ac]

sin C/2 = √[(s-a) (s-b)/ab]

cos A/2 = √s(s - a)/bc


cos B/2 = √s(s - b)/ac

cos C/2 = √s(s - c)/ab

tan A/2 = √[(s - b) (s - c)/s(s - a)]

tan B/2 = √[(s - c) (s - a)/s(s - b)]

tan C/2 = √[(s - a) (s - b)/s(s - c)]

Projection rule: In any ΔABC,
a = b cos C + c cos B

b = c cos A + a cos C

c = a cos B + b cos A

Area of a triangle
If Δ denotes the area of the triangle ABC, then it can be calculated in any of the following forms:

Δ = 1/2 bc sin A = 1/2 ca sin B = 1/2 ab sin C

Δ = √s(s - a)(s – b)(s - c)

Δ = 1/2. (a2 sin B sin C)/ sin(B + C)

= 1/2. (b2 sin C sin A)/ sin (C + A)

= 1/2. (c2 sin A sin B)/ sin (A + B)

Semi-perimeter of the triangle
If S denotes the perimeter of the triangle ABC, then s = (a + b + c)/2


Napier’s analogy
In any ΔABC,

tan [(B – C)/2] = (b – c)/(b + c) cot A/2

tan [(C – A) /2] = (c – a)/(c + a) cot B/2

tan [(A – B) /2] = (a – b)/(a + b) cot C/2

m-n theorem
Consider a triangle ABC where D is a point on side BC such that it divides the side BC in the ratio m: n, then as shown in the figure, the following results hold good:
Triangle ABC(m + n) cot θ = m cot α – n cot ß.

(m + n) cot θ = n cot B – m cot C.

Apollonius theorem
In a triangle ABC, if AD is the median through A, then

AB2 + AC2 = 2(AD2 + BD2).

If the three sides say a, b and c of a triangle are given, then angle A is obtained with the help of the formula
tan A/2 = √(s - b) (s - c) / s(s - a) or cos A = b2 + c2 - a2 / 2bc.

Angles B and C can also be obtained in the same way.

If two sides b and c and the included angle A are given, then
tan (B – C) /2
= (b – c)/ (b + c) cot A/2

This gives the value of (B- C)/2.

Hence, using (B + C)/2
= 90o - A/2 along with the last equation both B and C can be evaluated. Now, the sides can be evaluated using the formula

a = b sin A/sin B or a2 = b2 + c2 – 2bc cosA.

If two sides b and c and the angle B (opposite to side b) are given, then using the following results, we can easily obtain the remaining elements
sin C = c/b sinB, A = 180o – (B + C) and b = b sin A/sinB
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✅Notes on s-Block Elements:

Covalent Character:.

Small cation and large anion favors covalency.

Order: LiCl > NaCl > KCl > RbCl > CsCl & . LiI > LiBr > LiCl > LiF

Greater the charge on the cation greater is its polarizing power and hence larger is the covalent character: Na+CI- < Mg+2CI2 < AI+3 CI3

Greater the charge on the anion, more easily it gets polarized thereby imparting more covalent character to the compound formed eg covalent character increase in the order. NaCI < Na2SO4 < Na3PO4

c) Lattice Energies: Amount of energy required to separate one mole of solid ionic compound into its gaseous ions.

Greater the lattice energy, higher is the melting point of the alkali metals halide and lower is its solubility in water

d) Hydration Energy: Amount of energy released when one mole of gaseous ions combine with water to form hydrated ions.

M+ (g) + aq → M+ (aq) + hydration energy

X- (g) + aq → X- (aq) + hydration energy

Higher the hydration energy of the ions greater is the solubility of the compound in water.

The solubility of the most of alkali metal halides except those of fluorides decreases on descending the group since the decrease in hydration energy is more than the corresponding decrease in the lattice energy.

Due to high hydration energy of Li+ ion, Lithium halides are soluble in water except LiF which is sparingly soluble due to its high lattice energy.

For the same alkali metal the melting point decreases in the order

fluoride > chloride > bromide > iodide

For the same halide ion, the melting point of lithium halides are lower than those of the corresponding sodium halides and thereafter they decrease as we move down the group from Na to Cs.

The low melting point of LiCl (887 K) as compared to NaCl is probably because LiCl is covalent in nature and NaCl is ionic.

Anomalous Behavior of Lithium and diagonal relationship with Magnesium:

Li has anomalous properties due to

Very small size

High polarizing Power

Lithium show diagonal relationship with magnesium because both elements have almost same polarizing power.

The melting point and boiling point of lithium are comparatively high.

Lithium is much harder than the other alkali metals. Magnesium is also hard metal.

Lithium reacts with oxygen least readily to form normal oxide whereas other alkali metals form peroxides and superoxides.

LiOH like Mg (OH)2 is weak base. Hydroxides of other alkali metals are strong bases.

Due to their appreciable covalent nature, the halides and alkyls of lithum and magnesium are soluble in organic solvents.

Unlike elements of group 1 but like magnesium. Lithium forms nitride with nitrogen.6Li + N2 → 2Li3N

LiCl is deliquescent and crystallizes as a hydrate, LiCI2H2O. Other alkali metals do not form hydrates. also forms hydrate, MgCI2.8H2O .

Unlike other alkali metals lithium reacts directly with carbon to form an ionic carbide. Magnesium also forms a similar carbide.

The carbonates, hydroxides and nitrates of lithium as well as magnesium decompose on heating.
Li2CO3 → Li2O + CO2

MgCO3 → MgO + CO2

2LiOH → Li2O + H2O

Mg (OH)2 → MgO + H2O

4LiNO3 → 2Li2O + 4NO2 + O2

2Mg ( NO3)2 → 2Mg + 4NO2 +O2

The corresponding salts of other alkali metals are stable towards heat.
Lithium nitrate, on heating, decomposes to give lithium oxide, Li2O whereas other alkali metals nitrate decomposes to give the corresponding nitrite.

4LiNO3 → 2Li2O + 4NO2 + O2

2NaNO3 → 2NaNO2 + O2

2KNO3 → 2KNO2 + O2

Li2CO3, LiOH, LiF and Li3PO4 are the only alkali metal salts which are insoluble in water. The corresponding magnesium compounds are also insoluble in water.

Hydrogen carbonates of both lithium and magnesium can not be isolated in solid state. Hydrogen carbonates of other alkali metals can be isolated in solid state.

Sodium Hydroxide (NaOH):
NaOH is stable towards heat but is reduced to metal when heated with carbon

2NaOH + 2C → 2Na +2CO + H2

FeCl3 + 3NaOH →Fe(OH)3 + 3NaCl

NH4Cl + NaOH → NaCl + NH3 (pungent smell) + H2O
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Tricks for Conversions in Organic Chemistry

✅Haloalkanes can give you every possible functional group. They can also help you in increasing the chain size. So you might want to convert your starting material to haloalkanes and then go to the desired product

✅Remember the series of oxidation/reduction: Hydrocarbons (with various substituents) can be oxidised to alcohols, then appropriate carbonyl compounds and lastly carboxylic acids (or their derivatives)

✅Strong oxidizing agents are KMnO4 and K2Cr2O7

✅Mild oxidising agents depending on the situation are CrO4, Ammoniacal AgNO3, Benedicts Solution, Fehling’s solution, Cu or CuO at 573K, Bromine water etc.

✅Any carboxylic acid derivative can be converted back to the carboxylic acid by hydrolysis.

✅There are possibilities of hydride and methyl to get the most stable intermediate (carbocation/free radical)

✅A most common reaction, if your starting material is an alkane, is free-radical halogenation.

✅Saytzeff/Markovnikov's rules must be kept in mind while dealing with alkenes (they are based on the electron displacement effects only)

✅Decarboxylation and ozonolysis could go-to methods for decreasing the number of carbon atoms

✅Grignard reagent gives you the much-needed Alkyl nucleophile, which can be used at appropriate places.
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App Name : BooksBear 📚🐻
Category : E-Book Commerce

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