Question

1) х² + 12 х – 64 = 0 2) х² -11 х + 30 = 0 3) х² + 14 х + 24 = 0 4) 2х² + 3 = 3 - 7х 5) ( 3х – 1) (х + 2) = 20 РЕБЯЯТ УМОЛЯЮ КИДАЮ ВСЕ БАЛЛЫ

Answer 1

Ответ:

Объяснение:

1) $x^{2} + 12x - 64=0$

$D= 12^{2} - 4*1*(-64)= 400$

$x_{1} =\frac{-12+20 }{2}=4$

$x_{2} = \frac{-12-20}{2} = -16$

2) $x^{2} -11x+30=0$

$D= 11^{2} - 4*1*30=1$

$x_{1} = \frac{11+1}{2} =6\\x_{2} = \frac{11-1}{2} = 5$

3)

$x^{2} +14x +24=0\\D= 14^{2} - 4*1*24=100\\x_{1} = \frac{-14+10}{2} =-2\\x_{2} = \frac{-14-10}{2} =-12$

4) $2x^{2} +3=3-7x$

$2x^{2} =-7x\\2x^{2} +7x=0\\x*(2x+7)=0\\x_{1} =0 \\ x_{2} =\frac{-7}{2}$

5)$(3x-1)*(x+2)=20$

$3x^{2} +6x-x-2=20\\3x^{2} + 5x-2=20\\3x^{2} +5x-2-20=0\\3x^{2} +5x-22=0\\D=5^{2} -4*3*(-22)=289\\x_{1} = \frac{-5+17}{6} =2\\x_{2}= \frac{-5-17}{6} =\frac{-22}{6} = \frac{-11}{3}$

Answer 2

1. To solve the equation x² + 12x - 64 = 0, we can use the quadratic formula. The quadratic formula states that for an equation in the form ax² + bx + c = 0, the solutions for x can be found using the formula:

x = (-b ± √(b² - 4ac)) / (2a)

In this case, a = 1, b = 12, and c = -64. Plugging these values into the formula, we have:

x = (-12 ± √(12² - 4(1)(-64))) / (2(1))

Simplifying further:

x = (-12 ± √(144 + 256)) / 2 x = (-12 ± √400) / 2 x = (-12 ± 20) / 2

This gives us two possible solutions:

x1 = (-12 + 20) / 2 = 8/2 = 4 x2 = (-12 - 20) / 2 = -32/2 = -16

Therefore, the solutions to the equation x² + 12x - 64 = 0 are x = 4 and x = -16.

2. To solve the equation x² - 11x + 30 = 0, we can again use the quadratic formula. Comparing the equation to the form ax² + bx + c = 0, we have a = 1, b = -11, and c = 30. Applying the quadratic formula:

x = (-(-11) ± √((-11)² - 4(1)(30))) / (2(1)) x = (11 ± √(121 - 120)) / 2 x = (11 ± √1) / 2

Simplifying further:

x = (11 ± 1) / 2

This gives us two possible solutions:

x1 = (11 + 1) / 2 = 12/2 = 6 x2 = (11 - 1) / 2 = 10/2 = 5

Therefore, the solutions to the equation x² - 11x + 30 = 0 are x = 6 and x = 5.

3. To solve the equation x² + 14x + 24 = 0, we can once again use the quadratic formula. In this case, a = 1, b = 14, and c = 24. Applying the quadratic formula:

x = (-14 ± √(14² - 4(1)(24))) / (2(1)) x = (-14 ± √(196 - 96)) / 2 x = (-14 ± √100) / 2

Simplifying further:

x = (-14 ± 10) / 2

This gives us two possible solutions:

x1 = (-14 + 10) / 2 = -4/2 = -2 x2 = (-14 - 10) / 2 = -24/2 = -12

Therefore, the solutions to the equation x² + 14x + 24 = 0 are x = -2 and x = -12.

4. To solve the equation 2x² + 3 = 3 - 7x, we need to simplify it first.

By moving all terms to one side of the equation, we have:

2x² + 7x - 3 = 0

Now we can solve this quadratic equation using the quadratic formula. Here, a = 2, b = 7, and c = -3. Applying the quadratic formula:

x = (-7 ± √(7² - 4(2)(-3))) / (2(2)) x = (-7 ± √(49 + 24)) / 4 x = (-7 ± √73) / 4

So the solutions to the equation 2x² + 3 = 3 - 7x are:

x = (-7 + √73) / 4 x = (-7 - √73) / 4

5. To solve the equation (3x - 1)(x + 2) = 20, we can first expand the left side of the equation:

3x² + 6x - x - 2 = 20 3x² + 5x - 2 = 20

Now, we can move all terms to one side to form a quadratic equation:

3x² + 5x - 22 = 0

We can solve this quadratic equation using the quadratic formula. Here, a = 3, b = 5, and c = -22. Applying the quadratic formula:

x = (-5 ± √(5² - 4(3)(-22))) / (2(3)) x = (-5 ± √(25 + 264)) / 6 x = (-5 ± √289) / 6 x = (-5 ± 17) / 6

This gives us two possible solutions:

x1 = (-5 + 17) / 6 = 12/6 = 2 x2 = (-5 - 17) / 6 = -22/6 = -11/3

Therefore, the solutions to the equation (3x - 1)(x + 2) = 20 are x = 2 and x = -11/3.

I hope that helps! If you have any further questions, please let me know.