Question

what are the root of the quadratic equation
${x}^{2} = 9$
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Answer 1

Answer:

9x2 - 9(a + b)x + (2a2 + 4ab + ab + ab2) = 0 9x2 - (9a + 9b)x + [2a(a + 2b) + b(a + 2b)] = 0 9x2 - (9a + 9b)x + [(a + 2b)(2a + b)] = 0 9x2 - 3[(a + 2b) + (2a + b)]x + {(a + 2b)(2a + b)} = 0 9x2 - 3(a + 2b)x - 3(2a + b)x + {(a + 2b)(2a + b)} = 0 3x[3x - (a + 2b)] - (2a + b)[3x +(a - 2b)] = 0 [3x - (a + 2b)][3x -(2a + b)] = 0 [3x - (a + 2b)][3x - (2a + b)] = 0 [3x - (a + 2b)] = 0 or we can have [3x - (2a + b)] = 0 3x = (a + 2b) or 3x = (2a + b) Hence, x = a+2b3a+2b3 or x = 2a+b

Step-by-step explanation:

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Answer 2

❄️ Answer:-

x = 3 and -3

❄️ Solution:-

what are the root of the quadratic equation :

x² = 9

→ x² - 9 = 0

→ x² - 3² = 0

→ (x +3)(x - 3) = 0

→ x = 3 , -3

Therefore, the roots of the equation are 3 and (-3).