Question

2х^2+9х+с=0 х1=-5 х2-? с-?

Answer 1

Ответ:

Объяснение:

По теореме Виета

x₁+x₂=-9/2=-4.5

x₁x₂=c/2

x₂=-4,5-x₁=-4,5+5=0.5

c=2x₁x₂=2(-5)0,5=-5

Answer 2

To find the roots of the quadratic equation $2x^2 + 9x + c = 0$ when $x_1 = -5$, we can use the quadratic formula:

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

In your equation, $a = 2$, $b = 9$, and $c$ is the constant term we want to find.

Given that $x_1 = -5$, we can substitute this into the quadratic formula:

$-5 = \frac{-9 \pm \sqrt{9^2 - 4(2)(c)}}{2(2)}$

Now, we can solve for $c$. First, let's simplify the equation:

$-5 = \frac{-9 \pm \sqrt{81 - 8c}}{4}$

Now, we can isolate the term containing $c$:

$-5 \cdot 4 = -9 \pm \sqrt{81 - 8c}$

$-20 = -9 \pm \sqrt{81 - 8c}$

Next, add 9 to both sides:

$-20 + 9 = \pm \sqrt{81 - 8c}$

$-11 = \pm \sqrt{81 - 8c}$

Now, square both sides to eliminate the square root:

$(-11)^2 = (81 - 8c)$

$121 = 81 - 8c$

Now, solve for $c$:

$8c = 81 - 121$

$8c = -40$

$c = -40/8$

$c = -5$

So, the value of $c$ is -5.

Now, to find $x_2$, you can use the quadratic formula with the value of $c$ we just found:

$x_2 = \frac{-9 \pm \sqrt{9^2 - 4(2)(-5)}}{2(2)}$

Let's calculate $x_2$:

$x_2 = \frac{-9 \pm \sqrt{81 + 40}}{4}$

$x_2 = \frac{-9 \pm \sqrt{121}}{4}$

$x_2 = \frac{-9 \pm 11}{4}$

Now, you have two possibilities for $x_2$:

1. $x_2 = \frac{-9 + 11}{4} = \frac{2}{4} = \frac{1}{2}$

2. $x_2 = \frac{-9 - 11}{4} = \frac{-20}{4} = -5$

So, the two roots of the quadratic equation $2x^2 + 9x - 5 = 0$