Question

1) 6x2 + 5x – 1 = 02) 2x2 – 5x – 3 = 03) 2x2 – 7x + 5 = 0​

Answer 1

Пошаговое объяснение:

1) 6x^2+5x-1=0

По частному случаю теоремы Виета (a-b+c=0): 6-5-1=0

x1=-1

x2=-c/a=-(-1)/6=1/6

2) 2x^2-5x-3=0

D=b^2-4ac=(-5)^2-4*2*(-3)=25+24=7^2

x1=(-b+√D)/2a=(-(-5)+7)/2*2=12/4=3

x2=(-b-√D)/2a=(-(-5)-7)/2*2=-2/4=-0.5

3) 2x^2-7x+5=0

По частному случаю теоремы Виета (a+b+c=0): 2-7+5=0

x1=1

x2=c/a=5/2=2.5

Answer 2

These are quadratic equations, and you can solve them using the quadratic formula:

The quadratic formula is given by:

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

Where:

• $a$, $b$, and $c$ are the coefficients from the quadratic equation $ax^2 + bx + c = 0$.

Let's solve the equations one by one:

1. $6x^2 + 5x - 1 = 0$

In this equation, $a = 6$, $b = 5$, and $c = -1$.

Using the quadratic formula:

$x = \frac{-5 \pm \sqrt{5^2 - 4(6)(-1)}}{2(6)}$

$x = \frac{-5 \pm \sqrt{25 + 24}}{12}$

$x = \frac{-5 \pm \sqrt{49}}{12}$

$x = \frac{-5 \pm 7}{12}$

So, there are two solutions:

$x_1 = \frac{-5 + 7}{12} = \frac{2}{12} = \frac{1}{6}$

$x_2 = \frac{-5 - 7}{12} = \frac{-12}{12} = -1$

2. $2x^2 - 5x - 3 = 0$

In this equation, $a = 2$, $b = -5$, and $c = -3$.

Using the quadratic formula:

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

$x = \frac{5 \pm \sqrt{25 + 24}}{4}$

$x = \frac{5 \pm \sqrt{49}}{4}$

$x = \frac{5 \pm 7}{4}$

So, there are two solutions:

$x_1 = \frac{5 + 7}{4} = \frac{12}{4} = 3$

$x_2 = \frac{5 - 7}{4} = \frac{-2}{4} = -\frac{1}{2}$

3. $2x^2 - 7x + 5 = 0$

In this equation, $a = 2$, $b = -7$, and $c = 5$.

Using the quadratic formula:

$x = \frac{7 \pm \sqrt{(-7)^2 - 4(2)(5)}}{2(2)}$