Question

If the roots of the eqn (a2 + b2)x2 -2b(a+c)x+(b2 +c2) = 0

are equal, then--

a) 2b=a+c b) b2=ac c)b=2ac/a+c d) b=ac



Plz i want d answer solved... thnk u

Answer 1

Given:

The roots of the equation (a2 + b2)x2 -2b(a+c)x+(b2 +c2) = 0 are equal

To find:

If the roots of the eqn (a2 + b2)x2 -2b(a+c)x+(b2 +c2) = 0 are equal, then

Solution:  

From given, we have.

The roots of the eqn (a2 + b2)x2 -2b(a+c)x+(b2 +c2) = 0 are equal

The condition for the roots of a quadratic equation to be equal is,

b² - 4ac = 0

$\left(-2b\left(a+c\right)\right)^2-4\left(\left(a^2+b^2\right)\left(b^2+c^2\right)\right)=0\\\\-4b^4+8ab^2c-4a^2c^2=0$

$-4c^2a^2+8b^2ca-4b^4=0\\\\\left(8b^2c\right)^2-4\left(-4c^2\right)\left(-4b^4\right)=0\\\\a=\dfrac{-8b^2c}{2\left(-4c^2\right)}$

upon simplifying, we get,

-8ac² = -8b²c

ac² = b²c

ac = b²

Therefore, option B) b² = ac is correct option.

Answer 2

The roots of the equation (a² + b²)x² - 2b(a + c)x + (b² + c²) = 0 are equal.

To check : then

(a) 2b = a + c (b) b² = ac

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

solution : if a quadratic equation Ax² + Bx + C has equal roots. then Discriminant, D = B² - 4AC = 0

here in question, the roots of equation are equal.

so, D = {-2b(a + c)}² - 4(a² + b²)(b² + c²) = 0

⇒4b²(a + c)² - 4{a²(b² + c²) + b²(b² + c²)} = 0

⇒b²{a² + c² + 2ac} -{a²b² + a²c² + b⁴ + b²c²} = 0

⇒b²a² + b²c² + 2b²ac - a²b² - a²c² - b⁴ - b²c² = 0

⇒2b²ac - a²c² - b⁴ = 0

⇒b⁴ + a²c² - 2b²ac = 0

⇒(b²)² + (ac)² - 2(b²)(ac) = 0

⇒(b² - ac)² = 0

⇒b² = ac

Therefore the option (b) b² = ac, is correct choice.