19.1k views
4 votes
Hey can anyone pls answer these math problems!!

Hey can anyone pls answer these math problems!!-example-1

1 Answer

5 votes

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.

======================================================

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\\\\

======================================================

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.

======================================================

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\\\\

======================================================

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\\\\

======================================================

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)\\\\

User Cguedel
by
8.2k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories