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Quadratic Equations

1. Quadratic Equations

Quadratic Equations

Quadratic Polynomial:

**y = ax 2

• bx + c; a ≠ 0**

a = leading coefficient

b = coefficient of linear term

c = absolute term

**y = f(x) = ax 2

• bx + c**

In case

a = 0, b ≠ 0 ⇒ y = bx + c is linear polynomial

a = c = 0, b ≠ 0 ⇒ y = bx is odd linear polynomial

Cubic Polynomial:

y = ax^3 + bx^2 + cx + d; a ≠ 0

a = leading coefficient

d = absolute term

Roots of quadratic equation:

y = ax 2

• bx + c = 0

x =

2 b b 4ac

2a

− ± −

Where D = b 2

• 4ac is called discriminant.

Different graphs of Quadratic Expression:

**(i) Graph of y = ax 2

• bx + c; (a ≠ 0, a, b, c** ∈ R)

when a > 0, D < 0

a > 0 ⇒ Mouth facing upward

D < 0 ⇒ Parabola neither touch nor cut x– axis (no real root)

y > 0, ∀ x ∈ R

ax^2 + bx + c = 0

Sum of roots =

b

a

−

Product of roots =

c

a

D = b 2

• 4ac

3. Quadratic Equations

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**(iii) Graph of y = ax 2

• bx + c; (a** ≠ 0, a, b, c ∈ R) when a > 0 and D > 0

a > 0 ⇒ Mouth facing upward parabola. D > 0 ⇒ Parabola cuts the x–axis at 2

distinct points (two distinct real roots)

Plot y = x 2

• 3x + 2

D = 3^2 – 4(2) = 1 > 0

x 0 1 2 3 4 3/2 ∞ –∞

y 2 0 0 2 6 –1/4 ∞ ∞

y > 0 ⇒ x ∈ (– ∞ , 1) ∪ (2, ∞ )

y < 0 ⇒ x ∈ (1, 2)

y = 0 ⇒ x ∈ {1, 2}

(iv) Graph of y = ax 2 + bx + c; (a ≠ 0, a, b, c ∈ R)

when a < 0 and D < 0

a < 0 ⇒ Mouth facing downward

D < 0 ⇒ No real root

y < 0, ∀ x ∈ R

Plot y = –x^2 – 2x – 2

y = – (x + 1) 2

• 1

D < 0

Leading coefficient < 0

x 0 1 2 3 –1 –2 –3 ∞ –∞

y – 2 –5 –10 –17 –1 –2 –5 –∞ –∞

Quadratic Equations 4.

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**(v) Graph of y = ax 2

• bx + c; (a** ≠ 0, a, b, c ∈ R)

when a < 0 and D = 0

a < 0 ⇒ Mouth facing downward

D = 0 ⇒ Equal roots, i.e., parabola touches x–axis

y ≤ 0, ∀ x ∈ R

Plot y = – x^2 + 4x – 4

= – (x – 2) 2

D = 0

Leading coefficient < 0

x 0 1 2 3 4 –1 ∞ –∞

y –4 –1 0 – 1 –4 –9 –∞ –∞

(vi) Graph of y = ax^2 + bx + c; (a ≠ 0, a, b, c ∈ R)

when a < 0 and D > 0

a < 0 ⇒ Mouth facing downward

D > 0 ⇒ Two distinct real roots parabola cuts x– axis at two distinct points

Plot y = –x^2 + 3x – 2

y = – (x – 1) (x – 2)

D > 0

Leading coefficient < 0

x 0 1 2 3 4 –1 –2 ∞ –∞

y –2 0 0 –2 –6 –6 –12 –∞ –∞

Quadratic Equations 6.

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Both the roots of the equation (x – b) (x – c) + (x – c) (x – a) + (x – a) (x – b) = 0

are always

(A) Positive (B) Negative (C) Real (D) None of these

(C)

Given equation is 3x^2 – 2 (a+ b+ c)x + (ab + bc + ca) = 0

Now D = 4 (a + b + c)^2 –4 × 3 (ab + bc + ca)

= 2 {(a – b) 2

• (b – c) 2
• (c – a) 2 } Clearly, D ≥ 0 ⇒ both roots are always real.

The number of real solutions of the equation |x| 2

• 3|x| + 2 = 0 is (A) 4 (B) 1 (C) 3 (D) 2

(A)

Let |x| = t ∴ given equation is t^2 – 3t + 2 = 0

(t – 1) (t – 2) = 0 t = 1, t = 2

|x| = 1, |x| = 2

x = ± 1 , x = ± 2

Let f(x) be a quadratic expression which is positive for all real values of x. If g(x) = f(x) + f’(x) + f”(x) then for any real x :

(A) g(x) < 0 (B) g(x) > 0 (C) g(x) = 0 (D) g(x) ≥ 0

(B)

Let f(x) = ax^2 + bx + c (a ≠ 0, a, b, c ∈ R)

Also, f(x) > 0 ∀ x ∈ R ⇒ a > 0 and D < 0 Hence b^2 – 4ac < 0 …(i)

Now g(x) = (ax^2 + bx + c) + (2ax + b) + 2a = ax^2 + (b + 2a)x + (b + c + 2a)

D = (b + 2a) 2

• 4a (b + c + 2a) = b^2 + 4a^2 + 4ab – 4ab – 4ac – 8a^2

= b 2

• 4a 2 - 4ac = (b^2 – 4ac) – 4a^2 < 0 {from (i)}

Hence for g(x); D < 0, a > 0

⇒ g(x) > 0, ∀ x ∈ R

Let α^ , β be the roots of the equation (x – a) (x – b) = c, c ≠ 0. Then the roots of

the equation (x – α^ ) (x – β) + c = 0 are

(A) a, c (B) b, c (C) a, b (D) a + c, b + c

(C)

As α, β are roots of equation (x – a) (x – b) – c =

hence (x – a) (x – b) – c = (x – α) (x – β) ⇒ (x – a) (x – b) = (x – α) (x – β) + c

Clearly, roots of equation (x – α) (x – β) + c = 0 are a, b

Examples:

7. Quadratic Equations

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True/False If a < b < c < d then the roots of the equation

(x – a) (x – c) + 2(x – b) (x – d) = 0 are real and distinct

True

Let f(x) = (x – a) (x – c) + 2(x – b) (x – d)

Now f(a) = 2(a – b) (a – d) > 0 f(b) = (b – a) (b – c) < 0

f(c) = 2(c – b) (c – d) < 0 f(d) = (d – a) (d – c) > 0

Also, graph of f(x) is upward parabola.

Clearly, both roots of f(x) = 0 are real and distinct.

The number of points of intersection of two curves y = 2sinx and y = 5x^2 + 2x + 3 is

(A) 0 (B) 1 (C) 2 (D) ∞

A

y = 5x 2

• 2x + 3 D = 2^2 – 4(5) (3) = – 56 < 0

a = 5 > 0 ⇒ y > 0 ∀ x ∈R

Vertex

b D 1 14 , , 2a 4a 5 5

 −^ −^   −    =     

Clearly, y = 5x 2

• 2x + 3 ≥

14

5

and y = 2sinx ≤ 2

Hence, both curves do not intersect at any point.

For all x; x^2 + 2ax + 10 – 3a > 0 then interval in which a lies is:

(A) a < – 5 (B) – 5 < a < 2 (C) a > 5 (D) 2 < a < 5

B

D < 0

(2a) 2

• 4 (10 – 3a) < 0 4(a^2 – 10 + 3a) < 0

a 2

• 5a – 2a – 10 < 0 a(a + 5) – 2(a + 5) < 0

(a – 2) (a + 5) < 0 ⇒ a ∈ (–5, 2)

If b > a then the equation (x – a) (x– b) – 1 = 0 has

(A) Both roots in (a, b)

(B) Both roots in (– ∞ , a)

(C) both roots in (b, ∞ )

(D) one root in (– ∞ ,a) and the other in (b, ∞ )

9. Quadratic Equations

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If a, b ∈ R, a ≠ 0 and the quadratic equation ax 2

• bx + 1 = 0 has imaginary

roots then a + b+ 1 is (A) Positive (B) Negative

(C) zero (D) Depends on sign of b

A

∵ roots are imaginary of given equation.

∴ D < 0 ⇒ b 2

• 4a < 0

Now let f(x) = ax 2

• bx + 1

for which D < 0, f(0) = 1 > 0 Hence graph of f(x) is

⇒ f(x) > 0 ∀ x ∈ R

∴ f(– 1) > 0

a + b + 1 > 0

Multiple Objective Type

The graph of the quadratic polynomial; y = ax 2

• bx + c is as shown in the figure then:

(A) b 2

• 4ac > 0 (B) b < 0 (C) a > 0 (D) c < 0

ABCD

 Graph cuts x –axis at two distinct point

∴ D > 0 ⇒ b^2 – 4ac > 0

Graph is upward parabola

⇒ a > 0 Graph cuts negative y– axis

⇒ f(0) = c < 0

Also, vertex lies in IV quadrant

b 0 2a

−

b 0 2a

<

⇒ b < 0

Relation between roots and coefficient of quadratic equation

ax^2 + bx + c = 0; a ≠ 0, a, b, c ∈R

⇒ ax^2 + bx + c = a (x– α^ ) (x– β ) = 0

⇒ α^ + β =

− b

a

and αβ =

c

a

Know the facts

x^2 +

b c x a a

      +     

= (x – α^ ) (x – )

Quadratic Equations 10.

Formation of quadratic equation x^2 – (sum of roots)x + (product of roots) = 0

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Form a quadratic equation with rational coefficients whose one root is tan75°

 one root α^ = tan 75° = 2 + 3

∴ other root β = 2 – 3 as coefficients are rational.

Now α + (^) β = 4, (^) αβ = (^) ( )

2 2 2 − 3 = 1 required equation is

x 2

• ( α + (^) β ) x + αβ = 0

x 2

• 4x + 1 = 0

Form a quadratic equation with rational coefficients whose one root is cos 36 °

 one root α^ = cos 36° =

( 1 5 )

4

∴ other root

( 1 5 )

4

β = as coefficients are rational

Now α + (^) β =

1

2

, αβ =

( )

2 2 (^1 5 )

16 4

− = −

Required equation is:

x 2

• ( α + (^) β ) x + αβ = 0

2 1 1 x x 0 2 4

− − =

⇒ 4x^2 – 2x– 1=

Form a quadratic equation with rational coefficients whose one root is tan 8

π

 one root α^ = tan ( 1 ) 2

π = − +

∴ other root β =^ ( −^1 )^ −^2 as coefficient are rational

Now α + β = – 2, ( )

( )

2 2 αβ = − 1 − 2 = − 1

Required equation is:

2 x − α + β x + αβ = 0

x^2 + 2x – 1 = 0

Quadratic Equations 12.

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Find monic cubic polynomial with f(1) =1, f(2) =4, f(3) = 9

f(1) = 1^2 , f(2) = 2^2 , f(3) = 3^2

Then cubic is f(x) = (x – 1) (x – 2) (x – 3) + x 2

Solve (x – 7) (x – 3) (x + 5) (x + 1) = 1680

(x – 7) (x + 5) (x – 3) (x + 1) = 1680

(x 2

• 2x – 35) (x 2 - 2x – 3) = 1680

Let x 2

• 2x – 3 = α

Then ( α −^32 )α =^1680

α 2

• 32α – 1680 = 0

(α – 60) (α + 28) = 0

α – 60 = 0, α + 28 = 0

x 2

• 2x – 63 = 0 x 2 - 2x + 25 = 0

(x – 9) (x + 7) = 0 D < 0

x = – 7, 9

Solve (^) ( ) ( )

x^2 3 x^2 5 2 6 5 2 6 10

− −

• • − =

( 5 +^2 6 ) ( 5 −^2 6 )= 1

∴ (^) ( )

( )

2

2

x 3

x 3

−

−

= t (let)

Now equation t +

1

t

= 10 ⇒ t 2

• 10t + 1 = 0

t 5 2 6 2

t = 5 ± 2 6 ,

( ) ( ) ( ) ( )

x^2 3 x^2 3 5 2 6 5 2 6 , 5 2 6 5 2 6

− − −

• = + + = +

x 2

• 3 = 1, x 2 - 3 = – 1

x = ± 2, x = ± 2

x = {±√2, ± 2}

13. Quadratic Equations

Inequalities

Rule:

1. Adding positive number both sides inequality remains same.

Ex. 2 > 1 ⇒ 3 > 2

2. Subtracting both sides by positive number inequality remains same.

Ex. 2 > 1 ⇒ 1 > 0

3. Multiply and divide by positive number doesn’t affect inequality but

multiplying or dividing inequality with negative number changes sign of

inequality.

Ex. 4 > 2 ⇒ – 2 < – 1

Type – 1: Expression which can not be factorized

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x^2 + x + 1 > 0

 D = 1

2

• 4 (1) (1) < 0 and a = 1 > 0

∴ x 2

• x + 1 > 0, ∀ x ∈ R

∴ x ∈ R

x 2

• 3x + 4 < 0

D = (– 3)^2 – 4 (1) (4) < 0 and a = 1 > 0

∴ x^2 – 3x + 4 > 0, ∀ x ∈ R

∴ x ∈ ϕ

3. 3x^2 – 7x + 6 > 0

Sol. D = (–7) 2

• 4(3)(6) < 0 and a = 3 > 0 ∴ 3x^2 – 7x + 6 > 0, ∀ x ∈ R ∴ x ∈ R 4. – x 2
• 2x – 4 > 0

Sol. D = (–2)^2 – 4 (–1) (–4) < 0 and a = – 1 < 0

∴ – x 2

• 2x – 4 < 0, ∀ x ∈ R

∴ x ∈ ϕ

15. Quadratic Equations

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3x 2

• 7x + 4 ≥ 0

3x 2

• 3x – 4x + 4 ≥ 0

3x(x – 1) – 4 (x – 1) ≥ 0

(3x – 4) (x – 1) ≥ 0

( ]

4 x , 1 , 3

  ∈ −∞ ∪ ∞    

• •

• − + 4 3

1 x

Type–4: Repeated Linear factor

Rules:

(i) Factors with even power doesn’t affect sign.

(ii) Factors with odd power affect sign as linear.

(x + 1) (x – 3) (x – 2) 2

0

x ∈ (– ∞, – 1) ∪ (3, ∞)

x(x + 6) (x + 2) 2 (x – 3) > 0

x ∈ (– 6, – 2) ∪ (– 2, 0) ∪ (3, ∞)

(x – 1)^2 (x + 1)^3 (x – 4) < 0

x ∈ (– 1, 1) ∪ (1, 4)

Type–5: Rational Inequality

x 1 x 2 x 3 0 x 4 x 5

x ∈ ( −∞, 1 ) ∪ ( 2, 3 ) ∪( 4, 5)

2x 3 0 3x 7

− < −

2 x 3 / 2 0 3 x 7 / 3

x 3 / 2 0 x 7 / 3

3 7 x , 2 3

  ∈    

• • •

• (^) − − +

− 1 2 3 x

• • • •

− + (^) + − +

− 6 −^2 0 3 x

• • •

• (^) − − +

− 1 1 4 x

− • +^ • • +^ − • + 1 2 4 5 x

− (^) • 3

• •

• (^) − +

3

2

7

3

x

Quadratic Equations 16.

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2x 3 0 3x 7

− ≥ −

2 x 2 0 7 3 x 3

x 3 / 2 0 x 7 / 3

x , , 2 3

3 2 6

3 4

x 2x 3 x 4 0 x 3 3x 8

x 0, , 3 4 3 3

2

2

x 5x 12 3 x 4x 5

For x^2 – 4x +

D = 16 – 4(5) < 0, a = 1 > 0 ⇒ always positive Hence By cross multiplication

x^2 – 5x + 12 > 3x^2 – 12x + 15

2x 2

• 7x + 3 < 0 ⇒ 2x 2 - 6x – x + 3 < 0 (2x – 1) (x – 3) < 0

1 x , 3 2

  ∈    

2

2

x 5x 6 0 x x 1

− + <

For x^2 + x + 1

D = 1 2

• 4.1.1 < 0, a > 0 ⇒ always positive.

Hence given inequality reduces to x 2

• 5x + 6 < 0 (x – 2) (x – 3) < 0

X ∈ (2, 3)

• •

• (^) − +

3

2

7

3

x

• (^) • • − + 0 8

3

3 4 x

− (^) • − 3

2

• •
  • •

• (^) − +

1

2

3 x

• •

• (^) − +

2 3 x

Quadratic Equations 18.

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2 x 6x 7

x 4

• −

Clearly x + 4 > 0, ∀ x ∈ R − { − 4 }

Hence the inequality becomes 2 x + 6x − 7 < 0, x ≠ − 4

( x^ +^7 ) (^ x^ −^1 )<^ 0, x^ ≠ −^4

x ∈ (–7, 1) – {–4}

Let

x 1 x 3 y x 2

. Find the real values of x for which y takes real values.

y to be real

x 1 x 3 0 x 2

x ∈ [–1, 2) ∪ [3, ∞)

Find the set of all x for which 2

2x 1

2x 5x 2 x 1

≥

2

2x 1 0 2x 5x 2 x 1

− ≥

( ) (^) ( )

( ) (^ )

2

2

2x x 1 2x 5x 2 0 2x 5x 2 x 1

3x 2 0 2x 1 x 2 x 1

x 2, 1 , 3 2

 −^ −

Solve

2 x + 4x + 3 + 2x + 5 = 0

Consider ( ) ( )

2 x + 4x + 3 = x + 1 x + 3

Case– I:

Let

2 x + 4x + 3 ≥ 0 ⇒ x ∈ (^) ( −∞ −, (^3) ] ∪ −[ 1, ∞)

So, given equation becomes x^2 + 4x + 3 + 2x + 5 = 0

x 2

• 6x + 8 = 0

(x + 2) (x + 4) = 0

⇒ x = –2, x = –4 but x ∈ (–∞, –3] ∪ [–1, ∞) ⇒ x = –4 …(i)

• −

− 7 1

x

• •

− (^) • − +

− 1 2 3 x

• •

• (^) − − +

− 2 − 1 −^2 3

− 1 2

x

• •^ + •

• −

− (^3) − 1 x

• •

19. Quadratic Equations

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Case– II:

Let

2

x + 4x + 3 < 0 ⇒ x ∈ ( −3, − 1 )

then (^) ( )

2 equation becomes − x + 4x + 3 + 2x + 5 = 0

x 1 3 2

x = − 1 + 3 , x = − 1 − 3 but x ∈ ( −3, − 1 )

⇒ x = (^) ( − 1 − (^3) ) …(ii)

Now, (i) U (ii)

So, x = (^) { −4, − 1 − (^3) }

( ) ( )

2 2 x + 3x + 1 x + 3x − 3 ≥ 5

Let

2 x + 3x= α

2 α + 1 α − 3 − 5 ≥ 0 ⇒ α − 2 α − 8 ≥ 0

⇒ (α - 4) (α + 2) ≥ 0

( ) ( )

2 2 ⇒ x + 3x − 4 x + 3x + 2 ≥ 0

⇒ (x + 4) (x –1) (x + 1) (x + 2) ≥ 0

x ∈ (–∞, –4] ∪ [–2, –1] ∪ [1, ∞)

2

2

3x 7x 8 1 2 x 1

2 x + 1 > 0 ∀ x ∈R

∴ given inequality is 2 2 2 x + 1 < 3x − 7x + 8 ≤ 2x + 2

2 2 ⇒ x + 1 < 3x − 7x + 8 and 2 2 3x − 7x + 8 ≤ 2x + 2

2 2x − 7x + 7 > 0 and 2 x − 7x + 6 ≤ 0

2

2x − 7x + 7 > 0 and ( x − 1 ) ( x − 6 )≤ 0

(For 2x^2 –7x + 7 > 0; D < 0, a > 0) ∴ x ∈ R ∙ x ∈ [1,6]

x ∈ [1, 6]

• (^) − − +

− 4 −^2 −^1 1 x

• • + •