Question

1.4x^2-7х=0 2.-5х^2=0 3.4x^2-3x-1=0

Answer 1

Вот так как то короче .............

Answer 2

$1)4x^2-7x=0$
$x(4x-7)=0$
$x=0$ или  $4x-7=0$
                             $4x=7$
                             $x=7:4$
                             $x=1,75$
$Otvet:0;1,75.$
$2)-5x^2=0$
$x=0$
$Otvet:0.$
$3)4x^2-3x-1=0$
$D=b^2-4ac=(-3)^2-4*4*(-1)=9+16=25$
$\sqrt{D}=5$
$x_1= \frac{-b+ \sqrt{D} }{2a}= \frac{3+5}{8}=1$
$x_2= \frac{-b- \sqrt{D} }{2a}= \frac{3-5}{8}=-0,25$
$Otvet:1;-0,25.$

Answer 3

To solve these quadratic equations, we can use the quadratic formula:

For a quadratic equation in the form: ax^2 + bx + c = 0, the quadratic formula is:

x = (-b ± √(b^2 - 4ac)) / (2a)

Let's apply this formula to the equations you've provided:

1. 1.4x^2 - 7x = 0

In this equation, a = 1.4, b = -7, and c = 0. Plug these values into the quadratic formula:

x = (-(-7) ± √((-7)^2 - 4(1.4)(0))) / (2(1.4)) x = (7 ± √(49)) / (2.8) x = (7 ± 7) / 2.8

Now, we have two solutions:

x1 = (7 + 7) / 2.8 = 14 / 2.8 = 5 x2 = (7 - 7) / 2.8 = 0 / 2.8 = 0

So, the solutions for the first equation are x = 5 and x = 0.

2. -5x^2 = 0

In this equation, a = -5, b = 0, and c = 0. Plug these values into the quadratic formula:

x = (0 ± √(0^2 - 4(-5)(0))) / (2(-5)) x = (0 ± √(0)) / (-10) x = (0 ± 0) / (-10)

Since we have a square root of zero, there is only one solution:

x = 0 / (-10) = 0

So, the solution for the second equation is x = 0.

3. 4x^2 - 3x - 1 = 0

In this equation, a = 4, b = -3, and c = -1. Plug these values into the quadratic formula:

x = (-(-3) ± √((-3)^2 - 4(4)(-1))) / (2(4)) x = (3 ± √(9 + 16)) / 8 x = (3 ± √25) / 8 x = (3 ± 5) / 8

Now, we have two solutions:

x1 = (3 + 5) / 8 = 8 / 8 = 1 x2 = (3 - 5) / 8 = -2 / 8 = -1/4

So, the solutions for the third equation are x = 1 and x = -1/4.