Question

Number of distinct solution for equation (x - a)2 + (2 – 2a)2 + (x + 3a)2 = 0, if a is a real no. is​

Answer 1

Answer:

Distinct solution for equation (x - a)2 + (2 – 2a)2 + (x + 3a)2 = 0, if a is a real no. is​

Explanation:

The equation (x - a)2 + (2 – 2a)2 + (x + 3a)2 = 0 is a sum of squares of three expressions, x - a, 2 - 2a, and x + 3a.

For a given real number a, we can express each of the three expressions in terms of a and simplify the equation.

(x - a)2 + (2 – 2a)2 + (x + 3a)2 = (x^2 - 2ax + a^2) + (4 - 4a^2) + (x^2 + 6ax + 9a^2)

= x^2 - 2ax + a^2 + 4 - 4a^2 + x^2 + 6ax + 9a^2

= 2x^2 + 4ax + 14a^2

As we see that the equation is a quadratic equation in x, it will have two distinct solutions.

In summary, the equation (x - a)2 + (2 – 2a)2 + (x + 3a)2 = 0 for a real number a will have 2 distinct solutions.

It's worth noting that the equation (x - a)2 + (2 – 2a)2 + (x + 3a)2 = 0 is a sum of squares of three expressions, which are x - a, 2 - 2a, and x + 3a. As all the terms are squared, they are non-negative, so the equation is satisfied only when all three expressions are equal to zero.

However, since the expressions x - a, 2 - 2a, and x + 3a are not independent, we can't simultaneously set them to zero and get a solution. Therefore, the equation (x - a)2 + (2 – 2a)2 + (x + 3a)2 = 0 has no solution for any real value of a.

Another way to see this is by noting that the equation can be rewritten as (x - a + 2 - 2a + x + 3a)^2 = 0 which implies (2x + a)^2 = 0. This means that 2x + a = 0, so x = -(a/2) which is only possible if a = 0, but in that case, the equation is not true.

In conclusion, the equation (x - a)2 + (2 – 2a)2 + (x + 3a)2 = 0 is not a valid equation, and it has no solutions for any real value of a.

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Answer 2

$\frac{8a+\sqrt{-48a^{2} -4(8-16a) } }{4} ,\frac{8a-\sqrt{-48a^{2} -4(8-16a) } }{4}$.

• An equation is said to be linear if the maximum power of the variable is consistently 1. Another name for it is a one-degree equation. A linear equation with one variable has the conventional form Ax + B = 0. In this case, the variables x and A are variables, while B is a constant. A linear equation with two variables has the conventional form Ax + By = C.
• Each term in a linear equation has an exponent of 1, and when this algebraic equation is graphed, it always produces a straight line. It is called a "linear equation" for this reason.

Here, according to the given information, we are given that,

The equation given is of the form,

$(x-a)^{2} +(2-2a)^{2} +(x-3a)^{2}=0$

Or, $x^{2} -2xa+a^{2} +4-8a+4a^{2} +x^{2} -6xa+9a^{2} =0$

Or, $2x^{2} -8xa+14a^{2} +4-8a=0$

Then, by applying formula we get,

x = $\frac{-b+\sqrt{b^{2} -4ac} }{2a} ,\frac{-b-\sqrt{b^{2} -4ac} }{2a}$

Or, x = $\frac{8a+\sqrt{64a^{2} -4(28a^{2}+8-16a) } }{4} ,\frac{8a-\sqrt{64a^{2} -4(28a^{2}+8-16a) } }{4}$

Or, x = $\frac{8a+\sqrt{-48a^{2} -4(8-16a) } }{4} ,\frac{8a-\sqrt{-48a^{2} -4(8-16a) } }{4}$

Hence, the solutions are, $\frac{8a+\sqrt{-48a^{2} -4(8-16a) } }{4} ,\frac{8a-\sqrt{-48a^{2} -4(8-16a) } }{4}$, where a is a real number.

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