Question

1)х^2+x-4=02)4х^2+х+7=03)16х^2-40х+25=04)4х^2-55х+110√125=0

Answer 1

1)

X²+x-4=0

D=1+16=17

VD=V17

X1=(-1-V17)/2

X2=(-1+V17/2

2)

4x²+x +7=0

D=1-112 =-111   < 0  нет корни

3)

16x²-40x+25=0

D=1600-1600=0  равнение имеет двойной корен

X1,2=40/2*16=40/32=5/4

4)

4x²-55x+110V125=0

4x²-55x+500V5=0

D=3025-8000V5   < 0 нет корни

Answer 2

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

For an equation of the form ax^2 + bx + c = 0, the quadratic formula states that the solutions for x are given by:

x = (-b ± √(b^2 - 4ac)) / (2a)

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

1. x^2 + x - 4 = 0

a = 1, b = 1, c = -4

x = (-1 ± √(1^2 - 4(1)(-4))) / (2(1)) x = (-1 ± √(1 + 16)) / 2 x = (-1 ± √17) / 2

Therefore, the solutions are: x = (-1 + √17) / 2 x = (-1 - √17) / 2

2. 4x^2 + x + 7 = 0

a = 4, b = 1, c = 7

x = (-1 ± √(1^2 - 4(4)(7))) / (2(4)) x = (-1 ± √(1 - 112)) / 8 x = (-1 ± √(-111)) / 8

Since the discriminant is negative, the solutions are complex numbers: x = (-1 + i√111) / 8 x = (-1 - i√111) / 8

3. 16x^2 - 40x + 25 = 0

a = 16, b = -40, c = 25

x = (-(-40) ± √((-40)^2 - 4(16)(25))) / (2(16)) x = (40 ± √(1600 - 1600)) / 32 x = 40 / 32 x = 5 / 4

So the equation has one real solution: x = 5 / 4

4. 4x^2 - 55x + 110√125 = 0

a = 4, b = -55, c = 110√125

First, simplify the equation: 4x^2 - 55x + 110√125 = 0 4x^2 - 55x + 550√5 = 0

x = (-(-55) ± √((-55)^2 - 4(4)(550√5))) / (2(4)) x = (55 ± √(3025 - 8800√5)) / 8

The solutions will be complex numbers involving the square root of 5.