Question

solve this question very urgent

Answer 1

find the sum of the roots i. e -b/a and product of the roots i. e c/a and solve both

Answer 2

Answer:

The other two roots of the polynomial function is:

$\dfrac{7+3\sqrt{5}}{2},\dfrac{7-3\sqrt{5}}{2}$

Step-by-step explanation:

We are given a polynomial function as:

$f(x)=x^4-7x^3+10x^2-14x-2$

Now we are given two zeros of the function as:

$\sqrt{2}$ and $-\sqrt{2}$

Now let a and b be the other two zeros of he function.

We know that the sum of roots is given as:

$a+b+\sqrt{2}-\sqrt{2}=-\dfrac{-7}{1}\\\\a+b=7$

Similarly the product of roots is given as:

$a\times b\times \sqrt{2}\times (-\sqrt{2})=-2\\\\ab=1$

Hence,we have:

$b=\dfrac{1}{a}$

$a+\dfrac{1}{a}=7\\\\\\\dfrac{a^2+1}{a}=7\\\\\\a^2+1=7a\\\\a^2-7a+1=0$

Now the solution of quadratic equation is given as:

$a=\dfrac{7\pm \sqrt{49-4}}{2}\\\\a=\dfrac{7\pm 3\sqrt{5}}{2}$

Now when

$a=\dfrac{7+3\sqrt{5}}{2}$

Then we get:

$b=\dfrac{2}{7+3\sqrt{5}}\\\\=\dfrac{2}{7+3\sqrt{5}}\times \dfrac{7-3\sqrt{5}}{7-3\sqrt{5}}\\\\\\=\dfrac{2\times (7-3\sqrt{5})}{7^2-(3\sqrt{5})^2}\\\\\\=\dfrac{2\times (7-3\sqrt5)}{49-45}\\\\=\dfrac{2\times (7-3\sqrt{5})}{4}\\\\=\dfrac{7-3\sqrt{5}}{2}$

similarly when

$a=\dfrac{7-3\sqrt{5}}{2}$

we have:

$b=\dfrac{7+3\sqrt{5}}{2}$

Hence,

The other two roots of the polynomial function is:

$\dfrac{7+3\sqrt{5}}{2},\dfrac{7-3\sqrt{5}}{2}$