Question

solve the following quadratic equation by factorisation method.
solve all the parts

Answer 1

x2 - 2ax + a2 - b2 = 0
x2 - 2ax + (a+b)(a-b) = 0
x2 - 2ax + a + ba - b
sorry I can't do it

Answer 2

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Given,

3x – y = 3 and 9x – 3y = 9 are the two equations.

From 1st equation, we get,

x = (3+y)/3

Now, substitute the value of x in the given second equation to get,

9(3+y)/3 – 3y = 9

⇒9 +3y -3y = 9

⇒ 9 = 9

Therefore, y has infinite values and since, x = (3+y) /3, so x also has infinite values.

(iv) Given,

0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3are the two equations.

From 1st equation, we get,

x = (1.3- 0.3y)/0.2 _________________(1)

Now, substitute the value of x in the given second equation to get,

0.4(1.3-0.3y)/0.2 + 0.5y = 2.3

⇒2(1.3 – 0.3y) + 0.5y = 2.3

⇒ 2.6 – 0.6y + 0.5y = 2.3

⇒ 2.6 – 0.1 y = 2.3

⇒ 0.1 y = 0.3

⇒ y = 3

Now, substitute the value of y in equation (1), we get,

x = (1.3-0.3(3))/0.2 = (1.3-0.9)/0.2 = 0.4/0.2 = 2

Therefore, x = 2 and y = 3.