Sasi Math (Revised for 1024)

Scroll:Algebra new >> Root of the equation >> ps (4153)

Written Instructions:

For each Multiple Choice Question (MCQ), four options are given. One of them is the correct answer. Make your choice (1,2,3 or 4). Write your answers in the brackets provided..

For each Short Answer Question(SAQ) and Long Answer Question(LAQ), write your answers in the blanks provided.

Leave your answers in the simplest form or correct to two decimal places.

If the root of the equation (a² + b²)x² - 2(ac + bd)x + c2 + d2 = 0,

where ad - bc ≠ 0, are equal, prove that $\frac{a}{b}$ ≠ $\frac{c}{d}$

1) not equal
2) equal

Answer:_______________

If the root of the equation (a² + b²)x² - 2(ac + bd)x + c2 + d2 = 0,

where ad - bc ≠ 0, are equal, prove that $\frac{a}{b}$ ≠ $\frac{c}{d}$

1) equal
2) not equal

Answer:_______________

If the root of the equation (a² + b²)x² - 2(ac + bd)x + c2 + d2 = 0,

where ad - bc ≠ 0, are equal, prove that $\frac{a}{b}$ ≠ $\frac{c}{d}$

1) not equal
2) equal

Answer:_______________

If the root of the equation (a² + b²)x² - 2(ac + bd)x + c2 + d2 = 0,

where ad - bc ≠ 0, are equal, prove that $\frac{a}{b}$ ≠ $\frac{c}{d}$

1) equal
2) not equal

Answer:_______________

If the root of the equation (a² + b²)x² - 2(ac + bd)x + c2 + d2 = 0,

where ad - bc ≠ 0, are equal, prove that $\frac{a}{b}$ ≠ $\frac{c}{d}$

1) not equal
2) equal

Answer:_______________

If the root of the equation (a² + b²)x² - 2(ac + bd)x + c2 + d2 = 0,

where ad - bc ≠ 0, are equal, prove that $\frac{a}{b}$ ≠ $\frac{c}{d}$

1) equal
2) not equal

Answer:_______________

If the root of the equation (a² + b²)x² - 2(ac + bd)x + c2 + d2 = 0,

where ad - bc ≠ 0, are equal, prove that $\frac{a}{b}$ ≠ $\frac{c}{d}$

1) equal
2) not equal

Answer:_______________

If the root of the equation (a² + b²)x² - 2(ac + bd)x + c2 + d2 = 0,

where ad - bc ≠ 0, are equal, prove that $\frac{a}{b}$ ≠ $\frac{c}{d}$

1) not equal
2) equal

Answer:_______________

If the root of the equation (a² + b²)x² - 2(ac + bd)x + c2 + d2 = 0,

where ad - bc ≠ 0, are equal, prove that $\frac{a}{b}$ ≠ $\frac{c}{d}$

1) equal
2) not equal

Answer:_______________

10)

If the root of the equation (a² + b²)x² - 2(ac + bd)x + c2 + d2 = 0,

where ad - bc ≠ 0, are equal, prove that $\frac{a}{b}$ ≠ $\frac{c}{d}$

1) equal
2) not equal

Answer:_______________

If the root of the equation (a² + b²)x² - 2(ac + bd)x + c2 + d2 = 0,

where ad - bc ≠ 0, are equal, prove that $\frac{a}{b}$ ≠ $\frac{c}{d}$

1) not equal
2) equal

Answer: 2

SOLUTION 1 :

(a² + b²)x² - 2(ac + bd)x + c² + d² = 0

comparing with Ax² + Bx + C = 0

we get A = a² + b², B = -2(ac + bd) , C = c² + d²

given that the equation has equal roots.

in this â–º=0

B² - AC=0

[-2(ac + bd)]² - 4[(a² + b²) (c² + d²)] = 0

4(a²c² + 2abcd + b²d²) - 4(a²c² + a²d² +cb²c² +cb²d²) = 0

a²c²+ 2abcd + b²d² - a²c²- a²d² -b²c²- b²d²= 0

2abcd - a²d²- b²c² = 0

(ad - bc)² = 0

ad - bc = 0

ad = bc

$\frac{a}{b}$ = $\frac{c}{d}$

Hence proved.

If the root of the equation (a² + b²)x² - 2(ac + bd)x + c2 + d2 = 0,

where ad - bc ≠ 0, are equal, prove that $\frac{a}{b}$ ≠ $\frac{c}{d}$

1) equal
2) not equal

Answer: 1

SOLUTION 1 :

(a² + b²)x² - 2(ac + bd)x + c² + d² = 0

comparing with Ax² + Bx + C = 0

we get A = a² + b², B = -2(ac + bd) , C = c² + d²

given that the equation has equal roots.

in this â–º=0

B² - AC=0

[-2(ac + bd)]² - 4[(a² + b²) (c² + d²)] = 0

4(a²c² + 2abcd + b²d²) - 4(a²c² + a²d² +cb²c² +cb²d²) = 0

a²c²+ 2abcd + b²d² - a²c²- a²d² -b²c²- b²d²= 0

2abcd - a²d²- b²c² = 0

(ad - bc)² = 0

ad - bc = 0

ad = bc

$\frac{a}{b}$ = $\frac{c}{d}$

Hence proved.

If the root of the equation (a² + b²)x² - 2(ac + bd)x + c2 + d2 = 0,

where ad - bc ≠ 0, are equal, prove that $\frac{a}{b}$ ≠ $\frac{c}{d}$

1) not equal
2) equal

Answer: 2

SOLUTION 1 :

(a² + b²)x² - 2(ac + bd)x + c² + d² = 0

comparing with Ax² + Bx + C = 0

we get A = a² + b², B = -2(ac + bd) , C = c² + d²

given that the equation has equal roots.

in this â–º=0

B² - AC=0

[-2(ac + bd)]² - 4[(a² + b²) (c² + d²)] = 0

4(a²c² + 2abcd + b²d²) - 4(a²c² + a²d² +cb²c² +cb²d²) = 0

a²c²+ 2abcd + b²d² - a²c²- a²d² -b²c²- b²d²= 0

2abcd - a²d²- b²c² = 0

(ad - bc)² = 0

ad - bc = 0

ad = bc

$\frac{a}{b}$ = $\frac{c}{d}$

Hence proved.

If the root of the equation (a² + b²)x² - 2(ac + bd)x + c2 + d2 = 0,

where ad - bc ≠ 0, are equal, prove that $\frac{a}{b}$ ≠ $\frac{c}{d}$

1) not equal
2) equal

Answer: 2

SOLUTION 1 :

(a² + b²)x² - 2(ac + bd)x + c² + d² = 0

comparing with Ax² + Bx + C = 0

we get A = a² + b², B = -2(ac + bd) , C = c² + d²

given that the equation has equal roots.

in this â–º=0

B² - AC=0

[-2(ac + bd)]² - 4[(a² + b²) (c² + d²)] = 0

4(a²c² + 2abcd + b²d²) - 4(a²c² + a²d² +cb²c² +cb²d²) = 0

a²c²+ 2abcd + b²d² - a²c²- a²d² -b²c²- b²d²= 0

2abcd - a²d²- b²c² = 0

(ad - bc)² = 0

ad - bc = 0

ad = bc

$\frac{a}{b}$ = $\frac{c}{d}$

Hence proved.

If the root of the equation (a² + b²)x² - 2(ac + bd)x + c2 + d2 = 0,

where ad - bc ≠ 0, are equal, prove that $\frac{a}{b}$ ≠ $\frac{c}{d}$

1) not equal
2) equal

Answer: 2

SOLUTION 1 :

(a² + b²)x² - 2(ac + bd)x + c² + d² = 0

comparing with Ax² + Bx + C = 0

we get A = a² + b², B = -2(ac + bd) , C = c² + d²

given that the equation has equal roots.

in this â–º=0

B² - AC=0

[-2(ac + bd)]² - 4[(a² + b²) (c² + d²)] = 0

4(a²c² + 2abcd + b²d²) - 4(a²c² + a²d² +cb²c² +cb²d²) = 0

a²c²+ 2abcd + b²d² - a²c²- a²d² -b²c²- b²d²= 0

2abcd - a²d²- b²c² = 0

(ad - bc)² = 0

ad - bc = 0

ad = bc

$\frac{a}{b}$ = $\frac{c}{d}$

Hence proved.

If the root of the equation (a² + b²)x² - 2(ac + bd)x + c2 + d2 = 0,

where ad - bc ≠ 0, are equal, prove that $\frac{a}{b}$ ≠ $\frac{c}{d}$

1) not equal
2) equal

Answer: 2

SOLUTION 1 :

(a² + b²)x² - 2(ac + bd)x + c² + d² = 0

comparing with Ax² + Bx + C = 0

we get A = a² + b², B = -2(ac + bd) , C = c² + d²

given that the equation has equal roots.

in this â–º=0

B² - AC=0

[-2(ac + bd)]² - 4[(a² + b²) (c² + d²)] = 0

4(a²c² + 2abcd + b²d²) - 4(a²c² + a²d² +cb²c² +cb²d²) = 0

a²c²+ 2abcd + b²d² - a²c²- a²d² -b²c²- b²d²= 0

2abcd - a²d²- b²c² = 0

(ad - bc)² = 0

ad - bc = 0

ad = bc

$\frac{a}{b}$ = $\frac{c}{d}$

Hence proved.

If the root of the equation (a² + b²)x² - 2(ac + bd)x + c2 + d2 = 0,

where ad - bc ≠ 0, are equal, prove that $\frac{a}{b}$ ≠ $\frac{c}{d}$

1) not equal
2) equal

Answer: 2

SOLUTION 1 :

(a² + b²)x² - 2(ac + bd)x + c² + d² = 0

comparing with Ax² + Bx + C = 0

we get A = a² + b², B = -2(ac + bd) , C = c² + d²

given that the equation has equal roots.

in this â–º=0

B² - AC=0

[-2(ac + bd)]² - 4[(a² + b²) (c² + d²)] = 0

4(a²c² + 2abcd + b²d²) - 4(a²c² + a²d² +cb²c² +cb²d²) = 0

a²c²+ 2abcd + b²d² - a²c²- a²d² -b²c²- b²d²= 0

2abcd - a²d²- b²c² = 0

(ad - bc)² = 0

ad - bc = 0

ad = bc

$\frac{a}{b}$ = $\frac{c}{d}$

Hence proved.

If the root of the equation (a² + b²)x² - 2(ac + bd)x + c2 + d2 = 0,

where ad - bc ≠ 0, are equal, prove that $\frac{a}{b}$ ≠ $\frac{c}{d}$

1) equal
2) not equal

Answer: 1

SOLUTION 1 :

(a² + b²)x² - 2(ac + bd)x + c² + d² = 0

comparing with Ax² + Bx + C = 0

we get A = a² + b², B = -2(ac + bd) , C = c² + d²

given that the equation has equal roots.

in this â–º=0

B² - AC=0

[-2(ac + bd)]² - 4[(a² + b²) (c² + d²)] = 0

4(a²c² + 2abcd + b²d²) - 4(a²c² + a²d² +cb²c² +cb²d²) = 0

a²c²+ 2abcd + b²d² - a²c²- a²d² -b²c²- b²d²= 0

2abcd - a²d²- b²c² = 0

(ad - bc)² = 0

ad - bc = 0

ad = bc

$\frac{a}{b}$ = $\frac{c}{d}$

Hence proved.

If the root of the equation (a² + b²)x² - 2(ac + bd)x + c2 + d2 = 0,

where ad - bc ≠ 0, are equal, prove that $\frac{a}{b}$ ≠ $\frac{c}{d}$

1) equal
2) not equal

Answer: 1

SOLUTION 1 :

(a² + b²)x² - 2(ac + bd)x + c² + d² = 0

comparing with Ax² + Bx + C = 0

we get A = a² + b², B = -2(ac + bd) , C = c² + d²

given that the equation has equal roots.

in this â–º=0

B² - AC=0

[-2(ac + bd)]² - 4[(a² + b²) (c² + d²)] = 0

4(a²c² + 2abcd + b²d²) - 4(a²c² + a²d² +cb²c² +cb²d²) = 0

a²c²+ 2abcd + b²d² - a²c²- a²d² -b²c²- b²d²= 0

2abcd - a²d²- b²c² = 0

(ad - bc)² = 0

ad - bc = 0

ad = bc

$\frac{a}{b}$ = $\frac{c}{d}$

Hence proved.

10)

If the root of the equation (a² + b²)x² - 2(ac + bd)x + c2 + d2 = 0,

where ad - bc ≠ 0, are equal, prove that $\frac{a}{b}$ ≠ $\frac{c}{d}$

1) equal
2) not equal

Answer: 1

SOLUTION 1 :

(a² + b²)x² - 2(ac + bd)x + c² + d² = 0

comparing with Ax² + Bx + C = 0

we get A = a² + b², B = -2(ac + bd) , C = c² + d²

given that the equation has equal roots.

in this â–º=0

B² - AC=0

[-2(ac + bd)]² - 4[(a² + b²) (c² + d²)] = 0

4(a²c² + 2abcd + b²d²) - 4(a²c² + a²d² +cb²c² +cb²d²) = 0

a²c²+ 2abcd + b²d² - a²c²- a²d² -b²c²- b²d²= 0

2abcd - a²d²- b²c² = 0

(ad - bc)² = 0

ad - bc = 0

ad = bc

$\frac{a}{b}$ = $\frac{c}{d}$

Hence proved.