Question

(х^2+5х+1)(х^2+5х+3)=0

Answer 1

Ответ:

Объяснение:

Произведение равно нулю когда хотя бы один из множителей равен нулю

1) x²+5x+1=0

D= 25-4= 21

x1= (-5+√21)/2

x2= (-5-√21)/2

2) x²+5x+3=0

D= 25-12= 13

x3= (-5+√13)/2

x4= (-5-√13)/2

Answer 2

$(x^2+5x+1)(x^2+5x+3)=0\\\\\\ \bold{x^2+5x+1=0}\\D=5^2-4\cdot1\cdot1=25-4=21\\\sqrt{D}=\sqrt{21}\\x_1=\dfrac{-5+\sqrt{21}}{2}=\dfrac{\sqrt{21}-5}{2}\\x_2=\dfrac{-5-\sqrt{21}}{2}=-\dfrac{\sqrt{21}+5}{2}\\\\\\ \bold{x^2+5x+3=0}\\D=5^2-4\cdot1\cdot3=25-12=13\\\sqrt{D}=\sqrt{13}\\x_3=\dfrac{-5+\sqrt{13}}{2}=\dfrac{\sqrt{13}-5}{2}\\x_4=\dfrac{-5-\sqrt{13}}{2}=-\dfrac{\sqrt{13}+5}{2}$

ОТВЕТ:

$x_1=\dfrac{\sqrt{21}-5}{2}\\\\ x_2=-\dfrac{\sqrt{21}+5}{2}\\\\ x_3=\dfrac{\sqrt{13}-5}{2}\\\\ x_4=-\dfrac{\sqrt{13}+5}{2}$

Answer 3

To solve the quadratic equation $(x^2 + 5x + 1)(x^2 + 5x + 3) = 0$, you can factorize it and then solve for $x$.

Let's factorize each quadratic expression:

For the first quadratic $x^2 + 5x + 1$, the solutions can be found using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = 5$, and $c = 1$. Plugging in the values, we get:

$x = \frac{-5 \pm \sqrt{5^2 - 4(1)(1)}}{2(1)}$ $x = \frac{-5 \pm \sqrt{21}}{2}$

For the second quadratic $x^2 + 5x + 3$, you can similarly use the quadratic formula with $a = 1$, $b = 5$, and $c = 3$:

$x = \frac{-5 \pm \sqrt{5^2 - 4(1)(3)}}{2(1)}$ $x = \frac{-5 \pm \sqrt{13}}{2}$

Therefore, the solutions for the given equation $(x^2 + 5x + 1)(x^2 + 5x + 3) = 0$ are the combinations of these roots:

1. $x = \frac{-5 + \sqrt{21}}{2}$
2. $x = \frac{-5 - \sqrt{21}}{2}$
3. $x = \frac{-5 + \sqrt{13}}{2}$
4. $x = \frac{-5 - \sqrt{13}}{2}$