Question

1)Найдите все значения х, при которых равны значения выражений (х^2+1)^2 и 2x^2+2.2)Один из корней уравнения х^2+kx+45=0 равен 5. Найдите другой корень и коэффициент k.

Answer 1

1) $(x^2+1)^2=2x^2+2$

$x^4+2x^2+1-2x^2-2=0$

$x^4-1=0$ 

$(x-1)(x+1)(x^2+1)=0$ 

x = 1

x = -1

x = -i

x = i

 

2) $x^2+kx+45=0$ 

По теореме Виета, если х1, х2 - корни уранения ax^2+bx+c=0, то

$\left \{ {{x_1+x_2=-\frac{b}{a}} \atop {x_1x_2=\frac{c}{a}}} \right.$ 

 $\left \{ {{5+x_2=-k} \atop {5x_2=45}} \right.$ 

x2=9

5+9=-k

k = -14

Ответ: x^2-14x+45=0,   x1=5, x2=9 - корни 

 

Answer 2

I can help you with your math questions.

1) To find all the values of x for which the expressions $(x^2+1)^2$ and $2x^2+2$ are equal, we need to solve the following equation:

$$(x^2+1)^2=2x^2+2$$

Expanding the left side, we get:

$$x^4+2x^2+1=2x^2+2$$

Subtracting $2x^2+2$ from both sides, we get:

$$x^4=0$$

Taking the fourth root of both sides, we get:

$$x=\sqrt[4]{0}$$

Since the fourth root of zero is zero, we have:

$$x=0$$

Therefore, the only value of x that satisfies the equation is zero.

2) To find the other root and the coefficient k of the equation $x^2+kx+45=0$, given that one of the roots is 5, we can use the Vieta's formulas:

$$x_1+x_2=-k$$ $$x_1x_2=45$$

Substituting $x_1=5$, we get:

$$5+x_2=-k$$ $$5x_2=45$$

Solving for $x_2$, we get:

$$x_2=-\frac{k}{5}$$ $$x_2=9$$

Solving for k, we get:

$$k=-5-9$$ $$k=-14$$

Therefore, the other root is 9 and the coefficient k is -14.