Question

1)4х^2=9 2)7х^2-5х=0 3)2х^2-3х+5=0

Answer 1

$1) \: 4x {}^{2} = 9 \\ x {}^{2} = \frac{9}{4} \\ x _{1} = \sqrt \frac{9}{4} = \frac{3}{4} \\ x _{2} = - \sqrt \frac{9}{4} = - \frac{3}{4} \\Ответ: \: \frac{3}{4}; \: - \frac{3}{4} \\ 2) \: 7x {}^{2} - 5x = 0 \\ x(7x - 5) = 0 \\ x_{1} = 0 \\ 7x = 5 \\ x _{2} = \frac{5}{7}\\ Ответ: \:0; \: \frac{5}{7}\\ 3) \: 2x {}^{2} - 3x + 5 = 0 \\ D = 3 {}^{2} - 5 \times 2 \times 4 = 9 - 40 = - 31; D$

Answer 2

Ответ:

$4x {}^{2} = 9$

Х^2=9/4

Х=+-3/2

$7x {}^{2} - 5x = 0$

Х(7х-5)=0

Х=0 или 7х-5=0

7х=5

Х=5/7

$2x {}^{2} - 3x + 5 = 0$

D=b^2-4ac

D= (-3)^2 - 4*2*5= 9- 40= - 31

Дискриминант отрицательный, значит, решений нет.

Answer 3

Let's solve each of the given quadratic equations:

1. 4x^2 = 9

To solve this equation, we'll isolate x^2 by dividing both sides by 4:

4x^2/4 = 9/4

x^2 = 9/4

To find the values of x, we'll take the square root of both sides:

√(x^2) = ±√(9/4)

x = ±3/2

So the solutions to the equation 4x^2 = 9 are x = 3/2 and x = -3/2.

2. 7x^2 - 5x = 0

We can factor out x from the left side of the equation:

x(7x - 5) = 0

To find the solutions, we set each factor equal to zero:

x = 0 or 7x - 5 = 0

If 7x - 5 = 0, we can solve for x:

7x = 5 x = 5/7

Therefore, the solutions to the equation 7x^2 - 5x = 0 are x = 0 and x = 5/7.

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

This equation doesn't factor easily, so we'll use the quadratic formula to find the solutions:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this equation, a = 2, b = -3, and c = 5. Plugging these values into the quadratic formula:

x = (-(-3) ± √((-3)^2 - 4(2)(5))) / (2(2))

Simplifying further:

x = (3 ± √(9 - 40)) / 4

x = (3 ± √(-31)) / 4

Since the discriminant (the value inside the square root) is negative, there are no real solutions to this equation. The solutions would involve complex numbers.