Question

Find "a" for which roots of quadratic equation 2(a+5)x2-(a+5)x+1=0 are equal

Answer 1

EXPLANATION.

Quadratic equation.

⇒ 2(a + 5)x² - (a + 5)x + 1 = 0.

As we know that,

⇒ D = Discriminant Or b² - 4ac.

For equal roots : D = 0.

⇒ [-(a + 5)² - 4[2(a + 5)(1)] = 0.

⇒ (a + 5)² - 8(a + 5) = 0.

⇒ a² + 25 + 10a - 8a - 40 = 0.

⇒ a² + 2a - 15 = 0.

Factorizes the equation into middle term splits, we get.

⇒ a² + 5a - 3a - 15 = 0.

⇒ a(a + 5) - 3(a + 5) = 0.

⇒ (a - 3)(a + 5) = 0.

⇒ a = 3 and a = -5.

                                                                                                                     

MORE INFORMATION.

Nature of the roots of the quadratic expression.

(1) = Real and unequal, if b² - 4ac > 0.

(2) = Rational and different, if b² - 4ac is a perfect square.

(3) = Real and equal, if b² - 4ac = 0.

(4) = If D < 0 Roots are imaginary and unequal Or complex conjugate.

Answer 2

Step-by-step explanation:

EXPLANATION.

Quadratic equation.

⇒ 2(a + 5)x² - (a + 5)x + 1 = 0.

As we know that,

⇒ D = Discriminant Or b² - 4ac.

For equal roots : D = 0.

⇒ [-(a + 5)² - 4[2(a + 5)(1)] = 0.

⇒ (a + 5)² - 8(a + 5) = 0.

⇒ a² + 25 + 10a - 8a - 40 = 0.

⇒ a² + 2a - 15 = 0.

Factorizes the equation into middle term splits, we get.

⇒ a² + 5a - 3a - 15 = 0.

⇒ a(a + 5) - 3(a + 5) = 0.

⇒ (a - 3)(a + 5) = 0.

⇒ a = 3 and a = -5.

                                                                                                                     

MORE INFORMATION.

Nature of the roots of the quadratic expression.

(1) = Real and unequal, if b² - 4ac > 0.

(2) = Rational and different, if b² - 4ac is a perfect square.

(3) = Real and equal, if b² - 4ac = 0.

(4) = If D < 0 Roots are imaginary and unequal Or complex conjugate.

hope this helps you꧂

Thank U ✌︎✌︎