Question

plzz help
Math class 9​

Answer 1

Answer:

$8x - ay + {a}^{2} = 0 \\ put \: \: values \: \: of \: \: x \: \: and \: \: y \\( x = 1 \: \: and \: \: y \: = 6) \\ 8(1) - a(6) + {a}^{2} = 0 \\ 8 - 6a + {a}^{2} = 0 \\ {a}^{2} - 6a + 8 = 0 \\ {a}^{2} - (4a + 2a) + 8 = 0 \\ {a}^{2} - 4a - 2a + 8 = 0 \\ a(a - 4) - 2(a - 4) = 0 \\ (a - 4)(a - 2) = 0 \\ (a - 4) = 0 \: \: \: \: \: or \: \: \: \: \: (a - 2) = 0 \\ a = 4 \: \: \: \: \: or \: \: \: \: \: a = 2 \\ value \: \: of \: \: a \: \: can \: \: be \: \: 4 \: \: and \: \: 2$

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Answer 2

Answer:

x = 1, y = 6

8x - ay + a² = 0

=> 8*1 - a*6 + a² = 0

=> a² - 6a + 8 = 0

=> a² - 4a - 2a + 8 = 0

=> a (a - 4) - 2 (a - 4) = 0

=> (a - 4)(a - 2) = 0

if, a - 4 = 0

=> a = 4

or, a - 2 = 0

=> a = 2

The value of a is 4,2

Step-by-step explanation:

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