Quadratic equation lecture-1
Quadratic equation lecture-1
Quadratic equation
a x 2 + b x + c = 0 ; a , b , c ∈ C a x^2+b x+c=0 ; a, b, c \in \mathbb{C} a x 2 + b x + c = 0 ; a , b , c ∈ C a ≠ 0 a \neq 0 a = 0
If a ∈ R \quad a \in \mathbb{R} a ∈ R
W.L.O.G. Take ,a > 0 \quad a>0 a > 0
x = − b ± b 2 − 4 a c 2 a x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} x = 2 a − b ± b 2 − 4 a c
− b − b 2 − 4 a c 2 a = α , \frac{-b-\sqrt{b^2-4 a c}}{2 a}=\alpha, 2 a − b − b 2 − 4 a c = α ,
− b + b 2 − 4 a c 2 a = β \frac{-b+\sqrt{b^2-4 a c}}{2 a}=\beta 2 a − b + b 2 − 4 a c = β
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to →
Quadratic equation lecture-1
Discriminant of quadratic equation
a x 2 + b x + c = 0 a x^2+b x+c=0 a x 2 + b x + c = 0
⇔ a ( x 2 + b a x + c a ) = 0 \Leftrightarrow a\left(x^2+\frac{b}{a} x+\frac{c}{a}\right)=0 ⇔ a ( x 2 + a b x + a c ) = 0
⇔ a ( x 2 + 2 ⋅ x ⋅ b 2 a + b 2 4 a 2 − b 2 4 a 2 + c a ) \Leftrightarrow a\left(x^2+2 \cdot x \cdot \frac{b}{2 a}+\frac{b^2}{4 a^2}-\frac{b^2}{4 a^2}+\frac{c}{a}\right) ⇔ a ( x 2 + 2 ⋅ x ⋅ 2 a b + 4 a 2 b 2 − 4 a 2 b 2 + a c )
⇔ a { ( x + b 2 a ) 2 − ( b 2 − 4 a c 2 a ) 2 } = 0 \Leftrightarrow a\left\{\left(x+\frac{b}{2 a}\right)^2-\left(\frac{\sqrt{b^2-4 a c}}{2 a}\right)^2\right\}=0 ⇔ a { ( x + 2 a b ) 2 − ( 2 a b 2 − 4 a c ) 2 } = 0
D : = b 2 − 4 a c D:=b^2-4 a c D := b 2 − 4 a c
( Discriminant of the polynomial a x 2 + b x + c ) ( \text { Discriminant of the } \text { polynomial } a x^2+b x+c ) ( Discriminant of the polynomial a x 2 + b x + c )
Quadratic equation → \to → Discriminant of quadratic equation → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to →
Quadratic equation lecture-1
Nature of roots of quadratic equation
α = − b − D 2 a , β = − b + D 2 a \alpha=\frac{-b-\sqrt{D}}{2 a},\beta=\frac{-b+\sqrt{D}}{2a} α = 2 a − b − D , β = 2 a − b + D
If a , b , c ∈ Q a, b, c \in \mathbb{Q} a , b , c ∈ Q
D D D
⇒ α , β ∈ Q \Rightarrow \alpha, \beta \in \mathbb{Q} ⇒ α , β ∈ Q
D D D
⇒ α , β \Rightarrow \alpha, \beta ⇒ α , β
Quadratic equation → \to → → \to → Nature of roots of quadratic equation → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to →
Quadratic equation lecture-1
Nature of roots of quadratic equation
a , b , c ∈ R , a > 0 a, b, c \in \mathbb{R}, a>0 a , b , c ∈ R , a > 0
If D > 0 D>0 D > 0 α , β \alpha, \beta α , β
If D = 0 D=0 D = 0 α = β \alpha=\beta α = β
If D < 0 D<0 D < 0 α , β \alpha, \beta α , β
Quadratic equation → \to → → \to → Nature of roots of quadratic equation → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to →
Quadratic equation lecture-1
Case I
y = a x 2 + b x + c y=a x^2+b x+c y = a x 2 + b x + c a , b , c ∈ R a, b, c \in \mathbb{R} a , b , c ∈ R a > 0 a>0 a > 0
Case I: D > 0 D>0 D > 0
i) b < 0 , C > 0 b<0, C>0 b < 0 , C > 0
Quadratic equation → \to → → \to → → \to → Case I → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to →
Quadratic equation lecture-1
Case I
ii) b < 0 , c < 0 \text { ii) } b<0, c<0 ii) b < 0 , c < 0
iii) c < 0 , b > 0 \text {iii) } c<0, b>0 iii) c < 0 , b > 0
Quadratic equation → \to → → \to → → \to → Case I → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to →
Quadratic equation lecture-1
Case I
iv) c > 0 , b > 0 c>0, b>0 c > 0 , b > 0
Try {\text { Try }} Try
b = 0 or c = 0 {b=0} \quad \text { or } c=0 b = 0 or c = 0
Quadratic equation → \to → → \to → → \to → Case I → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to →
Quadratic equation lecture-1
Case II
Case II: D = 0 D = 0 D = 0
i) c > 0 , b > 0 c>0 , b>0 c > 0 , b > 0
ii) c > 0 , b > 0 c>0, b>0 c > 0 , b > 0
Quadratic equation → \to → → \to → → \to → → \to → Case II → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to →
Quadratic equation lecture-1
Case II
iii) c = 0 , b = 0 \text {iii) } c=0, b=0 iii) c = 0 , b = 0
D = b 2 − 4 a c = 0 D=b^2-4 a c=0 D = b 2 − 4 a c = 0
b 2 = 4 a c b^2=4 a c b 2 = 4 a c
a c ⩾ 0 a c \geqslant 0 a c ⩾ 0
c ⩾ 0 c \geqslant 0 c ⩾ 0
Quadratic equation → \to → → \to → → \to → → \to → Case II → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to →
Quadratic equation lecture-1
Case III
Case III: D < 0 D<0 D < 0
i) b < 0 , c > 0 \text { i) } b<0, c>0 i) b < 0 , c > 0
No real Solution α , β \alpha, \beta α , β
ii) \text{ii)} ii) b > 0 , c > 0 b>0, c>0 b > 0 , c > 0
No real solution
Quadratic equation → \to → → \to → → \to → → \to → → \to → Case III → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to →
Quadratic equation lecture-1
Case III
iii) b = 0 , c > 0 b=0, c>0 b = 0 , c > 0
No solution
D = b 2 − 4 a c < 0 D=b^2-4 a c<0 D = b 2 − 4 a c < 0
b 2 < 4 a c b^2<4 a c b 2 < 4 a c
⇒ c > b 2 4 a \Rightarrow c>\frac{b^2}{4 a} ⇒ c > 4 a b 2
b 2 ⩾ 0 , a > 0 b^2 \geqslant 0,a>0 b 2 ⩾ 0 , a > 0
⇒ b 2 4 a ⩾ 0 , c ⩾ 0 \Rightarrow \frac{b^2}{4 a} \geqslant 0, c \geqslant 0 ⇒ 4 a b 2 ⩾ 0 , c ⩾ 0
Quadratic equation → \to → → \to → → \to → → \to → → \to → Case III → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to →
Quadratic equation lecture-1
Sign of α , β \text{Sign of}\ \alpha, \beta Sign of α , β
a , b , c ∈ R a, b, c \in \mathbb{R} a , b , c ∈ R
For D < 0 D<0 D < 0
Consider D ⩾ 0 D \geqslant 0 D ⩾ 0
Case I: D = 0 , α = β = − b 2 a D=0, \quad \alpha=\beta=-\frac{b}{2 a} D = 0 , α = β = − 2 a b
i) if b > 0 b>0 b > 0 α , β < 0 \quad \alpha, \beta<0 α , β < 0
ii) if b < 0 b<0 b < 0 α , β > 0 \quad \alpha, \beta>0 α , β > 0
iii) if b = 0 b=0 b = 0 α = β = 0 \alpha=\beta=0 α = β = 0
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to →
Quadratic equation lecture-1
Sign of α , β \text{Sign of}\ \alpha, \beta Sign of α , β
Case II: D > 0 D>0 D > 0
− b ± D 2 a \frac{-b \pm \sqrt{D}}{2 a} 2 a − b ± D
a > 0 , 2 a > 0 a>0 , 2a>0 a > 0 , 2 a > 0
− b ± D -b \pm \sqrt{D} − b ± D
i) Let b > 0 b>0 b > 0
α = − b − b 2 − 4 a c 2 a < 0. \alpha=\frac{-b-\sqrt{b^2-4 a c}}{2 a}<0 . α = 2 a − b − b 2 − 4 a c < 0.
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to →
Quadratic equation lecture-1
Sign of α , β \text{Sign of}\ \alpha, \beta Sign of α , β
if c > 0 c>0 c > 0
$D =b^2-4 a c
$ \therefore \quad \sqrt{D}c$
∴ − b + D < 0 \therefore \quad-b+\sqrt{D}<0 ∴ − b + D < 0
∴ β = − b + D 2 a < 0 \therefore \quad \beta=\frac{-b+\sqrt{D}}{2 a}<0 ∴ β = 2 a − b + D < 0
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to →
Quadratic equation lecture-1
Sign of α , β \text{Sign of}\ \alpha, \beta Sign of α , β
if c = 0 \quad c=0 c = 0
β = 0 \beta=0 β = 0
β = − b + b 2 2 a = − b + b 2 a = 0 \beta =\frac{-b+\sqrt{b^2}}{2 a} =\frac{-b+b}{2 a}=0 β = 2 a − b + b 2 = 2 a − b + b = 0
If c < 0 \quad c<0\quad c < 0
D = b 2 − 4 a c > b 2 ∴ D > b D=b^2-4 a c>b^2 \qquad\therefore \sqrt{D}>b D = b 2 − 4 a c > b 2 ∴ D > b
∴ β = − b + D 2 a \therefore \quad \beta=\frac{-b+\sqrt{D}}{2 a} ∴ β = 2 a − b + D
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to →
Quadratic equation lecture-1
Sign of α , β \text{Sign of}\ \alpha, \beta Sign of α , β
ii) if b < 0 \quad b<0 b < 0
β = − b + D 2 a > 0
\beta=\frac{-b+\sqrt{D}}{2 a}>0
β = 2 a − b + D > 0
if c > 0 c>0 c > 0
$
D=b^2-4 a c
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to →
Quadratic equation lecture-1
Sign of α , β \text{Sign of}\ \alpha, \beta Sign of α , β
∴ D < − b
\therefore \sqrt{D}<-b ~
∴ D < − b D > 0 ~ D>0 D > 0
and b < 0 ~ b<0 b < 0
− b − D > 0 -b-\sqrt{D}>0 − b − D > 0
∴ α = − b − D 2 a > 0
\therefore \quad \alpha=\frac{-b-\sqrt{D}}{2 a}>0
∴ α = 2 a − b − D > 0
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to →
Quadratic equation lecture-1
Sign of α , β \text{Sign of}\ \alpha, \beta Sign of α , β
if c = 0 c=0 c = 0 α = 0 \alpha=0 α = 0
α = − b − b 2 2 a = − b + b 2 a = 0
\alpha = \frac{-b-\sqrt{b^2}}{2 a}=\frac{-b+b}{2 a}=0
α = 2 a − b − b 2 = 2 a − b + b = 0
if c < 0 , D > b 2 ∴ D > − b c<0, ~ ~ D>b^2 ~ ~ \therefore \sqrt{D}>-b c < 0 , D > b 2 ∴ D > − b
∴ α < 0 \therefore\alpha<0 ∴ α < 0
iii) if b = 0 , b=0,\quad b = 0 , ± D 2 a \frac{ \pm \sqrt{D}}{2 a} 2 a ± D
D > 0 , D>0,\quad D > 0 , α = − D 2 a < 0 \alpha=\frac{-\sqrt{D}}{2 a}<0 ~ ~ α = 2 a − D < 0 β = p 2 a > 0 ~ ~ \beta=\sqrt{\frac{p}{2 a}}>0 β = 2 a p > 0
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to →
Quadratic equation lecture-1
Recall
a x 2 + b x + c = 0 ; a , b , c ∈ C
a x^2+b x+c=0; \quad a, b, c \in \mathbb{C}
a x 2 + b x + c = 0 ; a , b , c ∈ C
α + β = − b a \alpha+\beta=-\frac{b}{a} α + β = − a b
2)α β = ( − b 2 a ) 2 − ( b 2 − 4 a c 2 a ) 2 = c a \alpha \beta =\left(-\frac{b}{2 a}\right)^2-\left(\frac{\sqrt{b^2-4 a c}}{2 a}\right)^2
=\frac{c}{a} α β = ( − 2 a b ) 2 − ( 2 a b 2 − 4 a c ) 2 = a c
∣ α − β ∣ = ∣ b 2 − 4 a c a ∣ |\alpha-\beta| ~ = ~ \mid \frac{\sqrt{b^2-4 a c}}{a}| ∣ α − β ∣ = ∣ a b 2 − 4 a c ∣
4)a α 2 + b α + c = 0 a\alpha^2+b \alpha+c=0 a α 2 + b α + c = 0
a β 2 + b β + c = 0 a\beta^2+b \beta+c=0 a β 2 + b β + c = 0
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → Recall → \to → → \to →
Quadratic equation lecture-1
Question
Let − π 6 < θ < − π 12 -\frac{\pi}{6}<\theta<-\frac{\pi}{12} − 6 π < θ < − 12 π α 1 , β 1 \alpha_1, \beta_1 α 1 , β 1 x 2 − 2 x sec θ + 1 = 0 x^2-2 x \sec \theta+1=0 x 2 − 2 x sec θ + 1 = 0 α 2 , β 2 \alpha_2, \beta_2 α 2 , β 2 x 2 + 2 x tan θ − 1 = 0 x^2+2 x \tan \theta-1=0 x 2 + 2 x tan θ − 1 = 0 α 1 > β 1 \alpha_1>\beta_1 α 1 > β 1 α 2 > β 2 \alpha_2>\beta_2 α 2 > β 2 α 1 + β 2 \alpha_1+\beta_2 α 1 + β 2
(1) 2 ( sec θ − tan θ ) 2(\sec \theta-\tan \theta) 2 ( sec θ − tan θ )
(2) 2 sec θ 2 \sec \theta 2 sec θ
(3) − 2 tan θ -2 \tan \theta − 2 tan θ
(4) 0 0 0
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → Question → \to →
Quadratic equation lecture-1
Solution
x 2 − 2 x sec θ + 1 = 0 x^2-2 x \sec \theta+1=0 x 2 − 2 x sec θ + 1 = 0
2 sec θ ± 4 sec 2 θ − 4 2 \frac{2 \sec \theta \pm \sqrt{4 \sec ^2 \theta-4}}{2} ~ 2 2 s e c θ ± 4 s e c 2 θ − 4 = sec θ ± tan θ ~ = \sec \theta \pm \tan \theta = sec θ ± tan θ
tan θ < 0 \tan \theta < 0 tan θ < 0
sec θ − tan θ ⏟ α 1 > sec θ + tan θ ⏟ β 2 \underset{\alpha_1}{\underbrace{\sec \theta-\tan \theta}}>\underset{\beta _2}{\underbrace{\sec \theta+\tan \theta}} α 1 sec θ − tan θ > β 2 sec θ + tan θ
x 2 + 2 × tan θ − 1 = 0 x^2+ 2 \times \tan \theta-1=0 x 2 + 2 × tan θ − 1 = 0
− 2 tan θ ± 4 tan 2 θ + 4 2 -\frac{2 \tan \theta \pm \sqrt{4 \tan ^2 \theta+4}}{2} − 2 2 t a n θ ± 4 t a n 2 θ + 4
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to → Solution
Quadratic equation lecture-1
Solution
= − tan θ ± sec θ = -\tan \theta \pm \sec \theta = − tan θ ± sec θ
[ sec θ > 0 ] [\sec \theta>0] [ sec θ > 0 ]
= − tan θ + sec θ ⏟ α 2 > − tan θ − sec θ ⏟ β 2 = ~ \underset{\alpha_2}{\underbrace{-\tan \theta+\sec \theta}}>\underset{\beta _2}{\underbrace{-\tan \theta -\sec \theta}} = α 2 − tan θ + sec θ > β 2 − tan θ − sec θ
α 1 + β 2 \alpha_1+\beta_2 α 1 + β 2
= sec θ − tan θ − tan θ − sec θ = \sec \theta-\tan \theta-\tan \theta-\sec \theta = sec θ − tan θ − tan θ − sec θ
= − 2 tan θ . =-2 \tan \theta . = − 2 tan θ .
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to → Solution
Quadratic equation lecture-1
Question
Let S = x ∈ R : x ≥ 0 S= x \in \mathbb{R}: x \geq 0 S = x ∈ R : x ≥ 0 2 ∣ x − 3 ∣ + x ( x − 6 ) + 6 = 0 2|\sqrt{x}-3|+\sqrt{x}(\sqrt{x}-6)+6=0 2∣ x − 3∣ + x ( x − 6 ) + 6 = 0 S S S
(1) contains exactly one element.
(2) contains exactly two elements.
(3) contains exactly four elements.
(4) is empty.
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → Question → \to →
Quadratic equation lecture-1
Solution
x − 3 ⩾ 0 when x ⩾ 9 \sqrt{x}-3 \geqslant 0 \text { when } x \geqslant 9 x − 3 ⩾ 0 when x ⩾ 9
x − 3 < 0 when x < 9 \sqrt{x}-3<0 \text { when } x<9 x − 3 < 0 when x < 9
[ x ⩾ 9 ] 2 ( x − 3 ) + ( x ) 2 − 6 x + 6 = 0 [x \geqslant 9] \quad 2(\sqrt{x}-3)+(\sqrt{x})^2-6 \sqrt{x}+6=0 [ x ⩾ 9 ] 2 ( x − 3 ) + ( x ) 2 − 6 x + 6 = 0
⇒ ( x ) 2 − 4 x = 0 \Rightarrow \quad(\sqrt{x})^2-4 \sqrt{x}=0 ⇒ ( x ) 2 − 4 x = 0
x ( x − 4 ) = 0 \sqrt{x}(\sqrt{x}-4)=0 x ( x − 4 ) = 0
x = 0 or x = 4 \sqrt{x}=0 \text { or } \sqrt{x}=4 x = 0 or x = 4
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to → Solution
Quadratic equation lecture-1
Solution
x = 0 x=0 x = 0 x = 16 x=16 x = 16
[ x < 9 ] − 2 ( x − 3 ) + ( x ) 2 − 6 x + 6 = 0 [x<9] \quad -2(\sqrt{x}-3)+(\sqrt{x})^2-6 \sqrt{x}+6=0 [ x < 9 ] − 2 ( x − 3 ) + ( x ) 2 − 6 x + 6 = 0
( x ) 2 − 8 x + 12 = 0 (\sqrt{x})^2-8 \sqrt{x}+12=0 ( x ) 2 − 8 x + 12 = 0
8 ± 64 − 48 2 \frac{ 8 \pm \sqrt{64-48}}{2} 2 8 ± 64 − 48
x = 2 or 6 \sqrt {x}=2 \quad \text { or }\quad 6 x = 2 or 6
x = 4 or x=4 \text { or } x = 4 or
x = 36 x=36 x = 36
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to → Solution
Quadratic equation lecture-1
Question
Number of real solutions of the equation ∣ x 2 + 4 x + 3 ∣ + 2 x + 5 = 0 \left|x^2+4 x+3\right|+2 x+5=0 x 2 + 4 x + 3 + 2 x + 5 = 0
(1) 0
(2) 1
(3) 2
(4) 4
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → Question → \to →
Quadratic equation lecture-1
Solution
x 2 + 4 x + 3 ⩾ 0 x^2+4 x+3 \geqslant 0 x 2 + 4 x + 3 ⩾ 0
x 2 + 6 x + 8 = 0 x^2+6 x+8=0 x 2 + 6 x + 8 = 0
x = − 6 ± 36 − 32 2 x =\frac{-6 \pm \sqrt{36-32}}{2} x = 2 − 6 ± 36 − 32
= − 6 ± 2 2 =\frac{-6 \pm 2}{2} = 2 − 6 ± 2
x = − 2 x = - 2 x = − 2 − 4 -4 − 4
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to → Solution
Quadratic equation lecture-1
Solution
x 2 + 4 x + 3 ⩾ 0 x^2+4 x+3 \geqslant 0 x 2 + 4 x + 3 ⩾ 0 [ x = − 2 ] \qquad[x=-2] [ x = − 2 ]
4 − 8 + 3 = − 1 < 0 4-8+3=-1<0\qquad 4 − 8 + 3 = − 1 < 0 [ x = − 4 ] [x=-4] [ x = − 4 ]
16 − 16 + 3 = 3 ⩾ 0 16-16+3=3 \geqslant 0 16 − 16 + 3 = 3 ⩾ 0
x 2 + 4 x + 3 < 0 x^2+4 x+3<0 x 2 + 4 x + 3 < 0
− x 2 − 4 x − 3 + 2 x + 5 = 0 -x^2-4 x-3+2 x+5=0 − x 2 − 4 x − 3 + 2 x + 5 = 0
x 2 + 2 x − 2 = 0 x^2+2 x-2=0 x 2 + 2 x − 2 = 0
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to → Solution
Quadratic equation lecture-1
Solution
x = − 2 ± 4 + 8 2 x=\frac{-2 \pm \sqrt{4+8}}{2} x = 2 − 2 ± 4 + 8 = − 2 ± 2 3 2 =\frac{-2 \pm 2 \sqrt{3}}{2} = 2 − 2 ± 2 3
x = − 1 ± 3 x=-1 \pm \sqrt{3} x = − 1 ± 3
[ x = − 1 ± 3 ] [x=-1 \pm \sqrt{3}] [ x = − 1 ± 3 ]
x 2 + 4 x + 3 < 0 x^2+4 x+3<0 x 2 + 4 x + 3 < 0
[ x = 3 − 1 ] [x=\sqrt{3}-1] [ x = 3 − 1 ]
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to → Solution
Quadratic equation lecture-1
Solution
3 − 2 3 + 1 + 4 3 − 4 + 3 3-2 \sqrt{3}+1+4 \sqrt{3}-4+ 3 3 − 2 3 + 1 + 4 3 − 4 + 3 = 2 3 + 3 > 0 =2 \sqrt{3}+3>0 = 2 3 + 3 > 0
[ x = − 1 − 3 ] [x=-1-\sqrt{3}] [ x = − 1 − 3 ]
( 3 + 1 ) 2 − 4 ( 3 + 1 ) + 3 (\sqrt{3}+1)^2-4(\sqrt{3}+1)+3 ( 3 + 1 ) 2 − 4 ( 3 + 1 ) + 3
= 3 + 2 3 + 1 − 4 3 − 4 + 3 = 3+2 \sqrt{3}+1-4 \sqrt{3}-4+3 = 3 + 2 3 + 1 − 4 3 − 4 + 3 = 3 − 2 3 < 0 = 3-2 \sqrt{3}<0 = 3 − 2 3 < 0
as 12 > 9 \text { as } 12>9 as 12 > 9
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to → Solution
Quadratic equation lecture-1
Question
The equation e sin x − e − sin x − 4 = 0 e^{\sin x}-e^{-\sin x}-4=0 e s i n x − e − s i n x − 4 = 0
(1) no real solutions.
(2) one real solutions.
(3) two real solutions.
(4) infinitely many real solutions.
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → Question → \to →
Quadratic equation lecture-1
Solution
e sin x − e − sin x − 4 = 0 e^{\sin x}-e^{-\sin x}-4=0 e s i n x − e − s i n x − 4 = 0
e 2 sin x − 4 e sin x − 1 = 0 e^{2 \sin x}-4 e^{\sin x}-1=0 e 2 s i n x − 4 e s i n x − 1 = 0
y = e sin x y=e^{\sin x} y = e s i n x
y 2 − 4 y − 1 = 0 y^2-4 y-1=0 y 2 − 4 y − 1 = 0
y = 4 ± 16 + 4 2 y=\frac{4 \pm \sqrt{16+4}}{2} y = 2 4 ± 16 + 4 = 4 ± 2 5 2 =\frac{4 \pm 2 \sqrt{5}}{2} = 2 4 ± 2 5
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to → Solution
Quadratic equation lecture-1
Solution
= 2 ± 5 =2 \pm \sqrt{5} = 2 ± 5
y ≠ 2 − 5 y \neq 2-\sqrt{5} y = 2 − 5
y = e sin x y= e^{\sin x} y = e s i n x
2 − 5 < 0 2-\sqrt{5}<0\quad 2 − 5 < 0 or 4 < 5 \text { or} ~ ~ 4<5 or 4 < 5
x ∈ R x \in \mathbb{R} x ∈ R
sin x ∈ R \sin x \in \mathbb{R} \quad\quad sin x ∈ R ∴ e sin x > 0 \therefore ~ e^{\sin x}>0 ∴ e s i n x > 0
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to → Solution
Quadratic equation lecture-1
Solution
y = 2 + 5 y=2+\sqrt{5} y = 2 + 5
e sin x = 2 + 5 e^{\sin x}=2+\sqrt{5} e s i n x = 2 + 5
∴ sin x = log ( 2 + 5 ) \therefore ~ \sin x=\log (2+\sqrt{5}) ∴ sin x = log ( 2 + 5 )
2 + 5 > 4 > e 2+\sqrt{5}>4>e 2 + 5 > 4 > e
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to → Solution
Quadratic equation lecture-1
Solution
∴ log ( 2 + 5 ) > log e = 1 \therefore \log (2+\sqrt{5})>\log e=1 ∴ log ( 2 + 5 ) > log e = 1
∴ sin x > 1 \therefore \quad \sin x>1 ∴ sin x > 1
x ∈ R x \in \mathbb{R} x ∈ R
− 1 ⩽ sin x ⩽ 1 -1 \leqslant \sin x \leqslant 1 − 1 ⩽ sin x ⩽ 1
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to → Solution
Quadratic equation lecture-1
Question
Let p , q , r p, q, r p , q , r p x 2 + q x + r = 0 p x^2+q x+r=0 p x 2 + q x + r = 0
(1) all positive real p p p r r r
(2) no p p p r r r
(3) ∣ p r − 7 ∣ > 4 3 \left|\frac{p}{r}-7\right|>4 \sqrt{3} r p − 7 > 4 3
(4) ∣ r p − 7 ∣ ≥ 4 3 \left|\frac{r}{p}-7\right| \geq 4 \sqrt{3} p r − 7 ≥ 4 3
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → Question → \to →
Quadratic equation lecture-1
Solution
[ q = p + r 2 ] [q=\frac{p+r}{2}] [ q = 2 p + r ]
q 2 − 4 p r ⩾ 0 q^2-4 p r \geqslant 0 q 2 − 4 p r ⩾ 0
( p + r ) 2 − 16 p r ⩾ 0 (p+r)^2-16 p r \geqslant 0 ( p + r ) 2 − 16 p r ⩾ 0
p 2 − 14 p r + r 2 ⩾ 0 p^2-14 p r+r^2 \geqslant 0 p 2 − 14 p r + r 2 ⩾ 0
1 − 14 r p + ( r p ) 2 ⩾ 0 1-14 \frac{r}{p}+\left(\frac{r}{p}\right)^2 \geqslant 0 1 − 14 p r + ( p r ) 2 ⩾ 0
( r p ) 2 − 2 r p 7 + 49 ⩾ 48 \left(\frac{r}{p}\right)^2-2 \frac{r}{p} 7+49 \geqslant 48 ( p r ) 2 − 2 p r 7 + 49 ⩾ 48
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to → Solution
Quadratic equation lecture-1
Solution
⇒ ( r p − 7 ) 2 ⩾ 48 \Rightarrow\left(\frac{r}{p}-7\right)^2 \geqslant 48 ⇒ ( p r − 7 ) 2 ⩾ 48
∴ ∣ r p − 7 ∣ ⩾ 4 3 . \therefore\left|\frac{r}{p}-7\right| \geqslant 4 \sqrt{3} . ∴ p r − 7 ⩾ 4 3 .
r p ⩾ 7 + 4 3 \frac{r}{p} \geqslant 7+4 \sqrt{3} p r ⩾ 7 + 4 3
− r p + 7 ⩾ 4 3 ⇒ r p ⩽ 7 − 4 3 -\frac{r}{p}+7 \geqslant 4 \sqrt{3} \Rightarrow \frac{r}{p} \leqslant 7-4 \sqrt{3} − p r + 7 ⩾ 4 3 ⇒ p r ⩽ 7 − 4 3
∴ r p ∈ ( 0 , 7 − 4 3 ] ∪ [ 7 + 4 3 , ∞ ) . \therefore \frac{r}{p} \in(0,7-4 \sqrt{3}] \cup[7+4 \sqrt{3}, \infty) . ∴ p r ∈ ( 0 , 7 − 4 3 ] ∪ [ 7 + 4 3 , ∞ ) .
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to → Solution
Quadratic equation lecture-1
Solution
when r p ∈ ( 0 , 7 − 4 3 ] \frac{r}{p} \in(0,7-4 \sqrt{3}] p r ∈ ( 0 , 7 − 4 3 ]
then p r ∈ [ 7 + 4 3 , ∞ ) \frac{p}{r} \in[7+4 \sqrt{3}, \infty) r p ∈ [ 7 + 4 3 , ∞ )
when r p ∈ [ 7 + 4 3 , ∞ ) \frac{r}{p} \in[7+4 \sqrt{3}, \infty) p r ∈ [ 7 + 4 3 , ∞ )
then p r ∈ ( 0 , 7 − 4 3 ] \frac{p}{r} \in(0,7-4 \sqrt{3}] r p ∈ ( 0 , 7 − 4 3 ]
∣ r p − 7 ∣ ⩾ 4 3 \left|\frac{r}{p}-7\right| \geqslant 4 \sqrt{3} p r − 7 ⩾ 4 3 ⇔ ∣ p p − 7 ∣ ⩾ 4 3 . \Leftrightarrow\left|\frac{p}{p}-7\right| \geqslant 4 \sqrt{3} . ⇔ p p − 7 ⩾ 4 3 .
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to → Solution
Quadratic equation lecture-1
Question
Let a , b , c > 0 a, b, c>0 a , b , c > 0 λ ∈ R \lambda \in \mathbb{R} λ ∈ R x 2 + 2 ( a + b + c ) x + 3 λ ( a b + b c + c a ) = 0 x^2+2(a+b+c) x+3 \lambda(a b+b c+c a)=0 x 2 + 2 ( a + b + c ) x + 3 λ ( ab + b c + c a ) = 0
(1) λ < 4 3 \lambda<\frac{4}{3} λ < 3 4
(2) λ > 5 3 \lambda>\frac{5}{3} λ > 3 5
(3) λ ∈ ( 1 3 , 5 3 ) \lambda \in\left(\frac{1}{3}, \frac{5}{3}\right) λ ∈ ( 3 1 , 3 5 )
(4) λ ∈ ( 4 3 , 5 3 ) \lambda \in\left(\frac{4}{3}, \frac{5}{3}\right) λ ∈ ( 3 4 , 3 5 )
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → Question → \to →
Quadratic equation lecture-1
Solution
4 ( a + b + c ) 2 − 12 λ ( a b + b c + c a ) ⩾ 0 4(a+b+c)^2-12 \lambda(a b+b c+c a) \geqslant 0 4 ( a + b + c ) 2 − 12 λ ( ab + b c + c a ) ⩾ 0
a 2 + b 2 + c 2 + 2 ( a b + b c + c a ) − 3 λ ( a b + b c + c a ) ⩽ 0 a^2+b^2+c^2+2(a b+b c+c a)-3 \lambda(a b+b c+c a) \leqslant 0 a 2 + b 2 + c 2 + 2 ( ab + b c + c a ) − 3 λ ( ab + b c + c a ) ⩽ 0
∴ 3 λ − 2 ⩽ a 2 + b 2 + c 2 a b + b c + c a . \therefore \quad 3 \lambda-2 \leqslant \frac{a^2+b^2+c^2}{a b+b c+c a} . ∴ 3 λ − 2 ⩽ ab + b c + c a a 2 + b 2 + c 2 .
a + b > c a+b>c a + b > c
⇒ a > c − b \rArr a>c-b ⇒ a > c − b
⇒ b > c − a \rArr b>c-a ⇒ b > c − a
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to → Solution
Quadratic equation lecture-1
Solution
a + c > b a+c>b a + c > b
⇒ a > b − c \rArr a>b-c ⇒ a > b − c
⇒ c > b − a \rArr c>b-a ⇒ c > b − a
b + c > a b+c>a b + c > a
⇒ b > a − c \rArr b>a-c ⇒ b > a − c
⇒ e > a − b \rArr e>a-b ⇒ e > a − b
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to → Solution
Quadratic equation lecture-1
Solution
a > ∣ c − b ∣ a>|c-b| a > ∣ c − b ∣
b > ∣ c − a ∣ b>|c-a| b > ∣ c − a ∣
c > ∣ b − a ∣ c>|b-a| c > ∣ b − a ∣
a 2 > b 2 + c 2 − 2 b c a^2>b^2+c^2-2 b c a 2 > b 2 + c 2 − 2 b c
b 2 > a 2 + c 2 − 2 a c b^2>a^2+c^2-2 a c b 2 > a 2 + c 2 − 2 a c
c 2 > a 2 + b 2 − 2 a b c^2>a^2+b^2-2 a b c 2 > a 2 + b 2 − 2 ab
a 2 + b 2 + c 2 > 2 ( a 2 + b 2 + c 2 ) − 2 ( a b + b c + c a ) a^2+b^2+c^2>2(a^2+b^2+c^2)-2(a b+b c+c a) a 2 + b 2 + c 2 > 2 ( a 2 + b 2 + c 2 ) − 2 ( ab + b c + c a )
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to → Solution
Quadratic equation lecture-1
Solution
∴ a 2 + b 2 + c 2 a b + b c + c a < 2 \therefore \frac{a^2+b^2+c^2}{a b+b c+c a}<2 ∴ ab + b c + c a a 2 + b 2 + c 2 < 2
a 2 + b 2 + c 2 a b + b c + c a ⩾ 3 λ − 2 \frac{a^2+b^2+c^2}{a b+b c+c a}\geqslant 3 \lambda-2 ab + b c + c a a 2 + b 2 + c 2 ⩾ 3 λ − 2
3 λ − 2 < 2 3 \lambda-2 <2 3 λ − 2 < 2
∴ λ < 4 3 \therefore \quad \lambda <\frac{4}{3} ∴ λ < 3 4
λ = 0 \lambda=0 λ = 0
x 2 + 2 ( a + b + c ) x = 0 x^2+2(a+b+c) x=0 x 2 + 2 ( a + b + c ) x = 0
Quadratic equation → \to → → \to → → \to → → \to → → \to → → \to → Sign of α , β \text{Sign of}\\ \alpha, \beta Sign of α , β → \to → → \to → → \to → Solution
