Question

If the roots of the equation
(a² + b²) x² - 2 b (a + c) x + (b² +c²) = 0 are equal then a, b, c
are in
(a) G.P. (b) A.P. (c) H.P. (d) none of these

Answer 1

Answer:

a) G.P.

Note:

• The general form of a quadratic equation is :

Ax² + Bx² + C = 0.

• The discriminant of the quadratic equation

Ax² + Bx + C = 0 is given as : D = B² - 4•A•C

• If D < 0 , then the roots are imaginary.

• If D = 0 , then the roots are real and equal.

• If D > 0 , then the roots are real and distinct.

• A.P. (Arithmetic Progression) : A sequence is said to be in A.P. if the difference between conservative terms are equal.

• The general form of an A.P. is given as ;

a , a + d , a + 2d , .... , a + (n-1)d where a is the first term and d is the common difference of A.P.

• If a , b , c are in A.P. then

=> b - a = c - b

=> b + b = a + c

=> 2b = a + c

• G.P. (Geometric Progression) : A sequence is said to be in G.P. if the ratio between conservative terms are equal.

• The general form of G.P. is given as ;

a , a•r² , a•r³ , .... , a•r^(n-1) where a is the first term and r is the common ratio of G.P.

• If a , b , c are in G.P. then

=> b/a = c/b

=> b•b = a•c

=> b² = a•c .

• H.P. (Harmonic Progression) : A sequence is said to be in H.P. if the reciprocals of its terms form an A.P.

• The general form of H.P. is given as ;

1/a , 1/(a+d) , 1/(a+2d) , ..... , 1/{a + (n-1)d}.

• If a , b , c are in H.P. then 1/a , 1/b , 1/c are in A.P.

=> 1/b - 1/a = 1/c - 1/b

=> 1/b + 1/b = 1/c + 1/a

=> 2/b = (a+c)/ac

=> 2ac = b(a+c)

Solution:

Here,

The given quadratic equation is :

(a²+b²)x² - 2b(a+c)x + (b²+c²) = 0 ----(1)

Also,

It is given that, a , b , c are equal.

Thus , let a = b = c = k .

Now,

Putting a = k , b = k , c = k in eq-(1) , we get ;

=> (k²+k²)x² - 2k(k+k)x + (k²+k²) = 0

=> 2k²x² - 2k•2kx + 2k² = 0

=> 2k²x² - 4k²x + 2k² = 0

=> 2k²(x² - 2x + 1) = 0

=> x² - 2x + 1 = 0

=> x² - x - x + 1 = 0

=> x(x-1) - (x-1) = 0

=> (x-1)(x-1) = 0

=> x = 1 , 1 (equal roots)

We get that ,

The roots of the given quadratic equation are equal , thus its discriminant must be equal to zero.

Thus,

=> D = 0

=> [-2b(a+c)]² - 4(a²+b²)(b²+c²) = 0

=> 4b²(a²+c²+2ac) - 4(a²b²+a²c²+b⁴+b²c²) = 0

=> b²(a²+c²+2ac) - (a²b²+a²c²+b⁴+b²c²) = 0

=> b²a²+b²c²+2acb²-a²b²-a²c²-b⁴-b²c² = 0

=> 2acb² - a²c² - b⁴ = 0

=> a²c² - 2acb² + b⁴ = 0

=> (ac)² - 2•ac•b² + (b²)² = 0

=> (ac - b²)² = 0

=> ac - b² = 0

=> b² = ac

Since,

b² = ac , thus a , b , c are in G.P.

Hence,

The required answer is ; a). G.P.

Answer 2

$\huge{\boxed{\red{\star\;Answer\;by\;Ra1}}}$

• Given , equation (a²+b²)x² -2b(a+c)x + (b²+c²) = 0

$\large{\boxed{\blue{\star\;A\;Note\;by\;Ra1}}}$

• If roots are equal then Discriminant , D = 0

Discriminent , D = b² - 4ac = 0

From the given equation ,

• a = (a²+b²)
• b = -2b(a+c)
• c = (b²+c²)

D = b² - 4ac = 0

(-2b(a+c))² -4(a²+b²)(b²+c²) = 0

4b²(a+c)² = 4(ab)² + 4(bc)² + 4(ac)² + 4(b)⁴

Cancelling 4 on both sides,

b²(a²+c²+2ac) = (ab)² + (bc)² + (ac)² + (b)⁴

(ab)² + (bc)² + 2ab²c = (ab)² + (bc)² + (ac)² + (b)⁴

2ab²c = (ac)² + b⁴

(ac)² + b⁴ - 2ab²c = 0

(ac - b²)² = 0

b² = ac

$\large{\boxed{\green{\star\;According\;to\;Ra1\;the\;required\;answer\;is\;G.P}}}$

$\huge{\boxed{\red{\star\;Dashanand\;Ravan\;hu\;mei}}}$