Question

please solve the question no 1,2 and 3 please

Answer 1

Answer:

нєу!!

Step-by-step explanation:

1. The perpendicular distance is 4.5 .

2. For a quadratic equation ax² + bx + c =0, the term b² - 4ac is called discriminant (D) of the quadratic equation because it  determines  whether the quadratic equation has real roots or not ( nature of roots).

D=  b² - 4ac

So a quadratic equation ax² + bx + c =0, has

i) Two distinct real roots, if b² - 4ac >0 , then x= -b/2a + √D/2a  &x= -b/2a - √D/2a

ii) Two equal real roots, if b² - 4ac = 0 , then x= -b/2a or -b/2a

iii) No real roots, if b² - 4ac <0

SOLUTION:

Given: 3x² – 4√3x + 4 = 0

On Comparing it with ax² + bx + c = 0, we get

a = 3, b = -4√3 and c = 4

Discriminant(D) = b² – 4ac

D= (-4√3)² – 4(3)(4)

D= 16 × 3 - 48

48 – 48 = 0

As , b² – 4ac = 0,

Hence,the given quadratic equation has  real and equal roots

3. sec²@ – tan²@ = 1

(3x)² – (3/x)² = 1

9x² – 9/x² = 1

9×(x² – 1/x²) = 1

Answer 2

hello meta...here is ur solution ☺️☺️



Step-by-step explanation:

1. The perpendicular distance is 4.5 .


2. For a quadratic equation ax² + bx + c =0, the term b² - 4ac is called discriminant (D) of the quadratic equation because it  determines  whether the quadratic equation has real roots or not ( nature of roots).

D=  b² - 4ac

So a quadratic equation ax² + bx + c =0, has

i) Two distinct real roots, if b² - 4ac >0 , then x= -b/2a + √D/2a  &x= -b/2a - √D/2a

ii) Two equal real roots, if b² - 4ac = 0 , then x= -b/2a or -b/2a

iii) No real roots, if b² - 4ac <0


SOLUTION:

Given: 3x² – 4√3x + 4 = 0

On Comparing it with ax² + bx + c = 0, we get

a = 3, b = -4√3 and c = 4


Discriminant(D) = b² – 4ac


D= (-4√3)² – 4(3)(4)

D= 16 × 3 - 48

48 – 48 = 0

As , b² – 4ac = 0,

Hence,the given quadratic equation has  real and equal roots

3. sec²@ – tan²@ = 1

(3x)² – (3/x)² = 1

9x² – 9/x² = 1

9×(x² – 1/x²) = 1

hope it helps u❤️❤️