Question

1)×(×-5)=24; 2)×(×+7)=78; 3)×(×+8)=105.

Answer 1

1) Х(Х-5)=24
Х^2-5X-24=0
X1*X2=-24
X1+X2=5
X1=-3
X2=8
2) x(x+7)=78
x^2+7x-78=0
x1*x2=-78
x1+x2=-7
x1=-13
x2=5
3)x(x+8)=105
x^2+8x-105=0
x1*x2=-105
x1+x2=-8
x1=7
x2=-15

Answer 2

1. 
x^2-5x-24=0
D=25-4*1*(-24)=25+96=121(корень=11)
x1=(5+11)\2=8
x2=(5-11)\2=-3
2.
x^2+7x-78=0
D=49-4*1*(-78)=49+312=361 (корень=19)
x1=(-7+19)\2=6
x2=(-7-19)\2=-13
3.
x^2+8x-105=0
D=64-4*1*(-105)=64+420=484 (корень=22)
x1=(-8+22)\2=7
x2=(-8-22)\2=-15

Answer 3

To solve these equations, we can use the method of completing the square.

1. ×(×-5)=24

Let's add (5/2)^2 = 6.25 to both sides of the equation to complete the square:

×(×-5) + 6.25 = 30.25

Now we can write the left-hand side as a square:

(× - 2.5)^2 = 30.25

Taking the square root of both sides, we get:

× - 2.5 = ±5.5

So × can be either:

× = 2.5 + 5.5 = 8 or × = 2.5 - 5.5 = -3

Therefore, the solutions are × = 8 and × = -3.

2. ×(×+7)=78

Again, let's add (7/2)^2 = 24.5 to both sides of the equation to complete the square:

×(×+7) + 24.5 = 102.5

Now we can write the left-hand side as a square:

(× + 3.5)^2 = 102.5

Taking the square root of both sides, we get:

× + 3.5 = ±√102.5

So × can be either:

× = -3.5 + √102.5 ≈ 6.47 or × = -3.5 - √102.5 ≈ -13.97

Therefore, the solutions are × ≈ 6.47 and × ≈ -13.97.

3. ×(×+8)=105

Once again, let's add (8/2)^2 = 16 to both sides of the equation to complete the square:

×(×+8) + 16 = 121

Now we can write the left-hand side as a square:

(× + 4)^2 = 121

Taking the square root of both sides, we get:

× + 4 = ±11

So × can be either:

× = -4 + 11 = 7 or × = -4 - 11 = -15

Therefore, the solutions are × = 7 and × = -15.