Question

11.
If the roots of the equation X2 - 8x + 22 - 6a = 0
are real and distinct, then find all possible
values of a​

Answer 1

Correct equation:-

• x² - 8x + (a² - 6a) = 0 .

Given:-

• The roots of the equation are real and distinct.

To find:-

• All possible values of a.

Solution:-

➤ For any  quadratic  equation having real roots , the discriminant must be greater than or equal to zero .

We know,

➤  [tex] \pink{b² - 4ac ≥ 0 }[/tex]

Now substituiting the given values :

=>  (8)² - 4(1)(a² - 6a)  ≥ 0

=> 64 - 4a² + 24a  ≥ 0

=> 4a² - 24a - 64 ≤ 0

=> a² - 6a - 16 ≤ 0

=> (a + 2) (a - 8) ≤ 0

Hence  the range:

➤  -2 ≤ a ≤ 8

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Answer:-

➤ $\boxed{\underline{\orange{A ∈ -2,8 }}}$

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Answer 2

Correct equation:-

x² - 8x + (a² - 6a) = 0

.

Given:-

The roots of the equation are real and distinct.

To find:-

All possible values of a.

Solution:-

➤ For any  quadratic  equation having real roots , the discriminant must be greater than or equal to zero

.

We know,

➤  $\pink{b² - 4ac ≥ 0</p><p>}$

Now substituiting the given values :

=>  (8)² - 4(1)(a² - 6a)  ≥ 0

=> 64 - 4a² + 24a  ≥ 0

=> 4a² - 24a - 64 ≤ 0

=> a² - 6a - 16 ≤ 0

=> (a + 2) (a - 8) ≤ 0

Hence  the range:

➤  -2 ≤ a ≤ 8

___________________

Answer:-

➤ $\boxed{\underline{\orange{A ∈ -2,8 }}}$

______________________________