Question

Hey can anyone pls answer these math problems!!

Answer 1

Problem 1

We have something in the form
`ax^2+bx+c = 0`, where

• a = 1
• b = -8
• c = 15

Those values are plugged into the quadratic formula as shown below.

`x = (-b\pm√(b^2-4ac))/(2a)\\\\x = (-(-8)\pm√((-8)^2-4(1)(15)))/(2(1))\\\\x = (8\pm√(4))/(2)\\\\x = (8\pm2)/(2)\\\\x = (8+2)/(2) \ \text{ or } \ x = (8-2)/(2)\\\\x = (10)/(2) \ \text{ or } \ x = (6)/(2)\\\\x = 5 \ \text{ or } \ x = 3\\\\`

The roots or solutions are x = 5 and x = 3

To check these answers, plug them back into the original equation and you should get the same number on both sides. You can also use a graphing calculator to confirm the answers. In this case, the parabola crosses the x axis at 5 and 3.

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Problem 2

This time we have

• a = 1
• b = 14
• c = 0

which leads to:

`x = (-b\pm√(b^2-4ac))/(2a)\\\\x = (-(14)\pm√((14)^2-4(1)(0)))/(2(1))\\\\x = (-14\pm√(196))/(2)\\\\x = (-14\pm14)/(2)\\\\x = (-14+14)/(2) \ \text{ or } \ x = (-14-14)/(2)\\\\x = (0)/(2) \ \text{ or } \ x = (-28)/(2)\\\\x = 0 \ \text{ or } \ x = -14\\\\`

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Problem 3

We'll plug in a = 1, b = -5, and c = 1

`x = (-b\pm√(b^2-4ac))/(2a)\\\\x = (-(-5)\pm√((-5)^2-4(1)(1)))/(2(1))\\\\x = (5\pm√(21))/(2)\\\\x = (5+√(21))/(2) \ \text{ or } \ x = (5-√(21))/(2)\\\\`

Unlike the previous results, we cannot simplify the square roots down to some whole number. So we leave the roots as they are.

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Problem 4

We first need to move the 12 over to the left side so that we have 0 on the right side.

The equation
`x^2-x = 12` becomes
`x^2-x-12 = 0` at which point we have

• a = 1
• b = -1
• c = -12

So,

`x = (-b\pm√(b^2-4ac))/(2a)\\\\x = (-(-1)\pm√((-1)^2-4(1)(-12)))/(2(1))\\\\x = (1\pm√(49))/(2)\\\\x = (1\pm7)/(2)\\\\x = (1+7)/(2) \ \text{ or } \ x = (1-7)/(2)\\\\x = (8)/(2) \ \text{ or } \ x = (-6)/(2)\\\\x = 4 \ \text{ or } \ x = -3\\\\`

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Problem 5

Start by subtracting 5 from each side

The equation
`x^2-10x+14 = 5` becomes
`x^2-10x+9 = 0`

We'll plug in a = 1, b = -10, c = 9

`x = (-b\pm√(b^2-4ac))/(2a)\\\\x = (-(-10)\pm√((-10)^2-4(1)(9)))/(2(1))\\\\x = (10\pm√(64))/(2)\\\\x = (10\pm8)/(2)\\\\x = (10+8)/(2) \ \text{ or } \ x = (10-8)/(2)\\\\x = (18)/(2) \ \text{ or } \ x = (2)/(2)\\\\x = 9 \ \text{ or } \ x = 1\\\\`

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Problem 6

This is similar to the first three problems where the equation is already in
`ax^2+bx+c = 0` form.

This time we have a = 1, b = 5, c = 3.

`x = (-b\pm√(b^2-4ac))/(2a)\\\\x = (-(5)\pm√((5)^2-4(1)(3)))/(2(1))\\\\x = (-5\pm√(13))/(2)\\\\x = (-5+√(13))/(2) \ \text{ or } \ (-5-√(13))/(2)\\\\`