Question

Сумма нулей функции у=(х-1/√3+√2)(х-√3-√2) равна​

Answer 1

$y=(x-(1/\sqrt{3}-\sqrt{2}))(x-(\sqrt{3}+\sqrt{2}))\\1/\sqrt{3}+\sqrt{3}\\\frac{4\sqrt{3} }{3}$

Answer 2

To find the sum of the zeros of the function $y = (x - \frac{1}{\sqrt{3}} + \sqrt{2})(x - \sqrt{3} - \sqrt{2})$, we need to determine the values of $x$ that make $y$ equal to zero. In other words, we need to find the roots of the function.

Setting the function equal to zero and solving for $x$:

$(x - \frac{1}{\sqrt{3}} + \sqrt{2})(x - \sqrt{3} - \sqrt{2}) = 0$

To solve this, we can use the zero product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. So, we have:

$x - \frac{1}{\sqrt{3}} + \sqrt{2} = 0 \quad \text{or} \quad x - \sqrt{3} - \sqrt{2} = 0$

Now, let's solve for $x$ in each case:

Case 1:

$x - \frac{1}{\sqrt{3}} + \sqrt{2} = 0$ $x = \frac{1}{\sqrt{3}} - \sqrt{2}$

Case 2:

$x - \sqrt{3} - \sqrt{2} = 0$ $x = \sqrt{3} + \sqrt{2}$

The sum of the zeros (roots) of the function is the sum of the values we found in both cases:

Sum of zeros = $\frac{1}{\sqrt{3}} - \sqrt{2} + \sqrt{3} + \sqrt{2}$

The term $\sqrt{2}$ cancels out, and we are left with:

Sum of zeros = $\frac{1}{\sqrt{3}} + \sqrt{3}$

To rationalize the denominator, we multiply the fraction by $\frac{\sqrt{3}}{\sqrt{3}}$:

Sum of zeros = $\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} + \sqrt{3} \cdot \frac{\sqrt{3}}{\sqrt{3}}$