Question

HELP HELP HELP

EASY 18 POINTS!


Solving Quadratic Equations Algebraically Explain how to determine the solutions to a quadratic equation, algebraically. Create a problem with a quadratic equation that you would solve algebraically and solve it.

Answer 1

To solve a quadratic equation algebraically, you can use the quadratic formula, substituting the coefficients of the equation into the formula. Let's create a problem and solve it algebraically. Problem: Solve the quadratic equation 2x^2 + 5x - 3 = 0 algebraically. Solution: Substitute the coefficients into the quadratic formula and simplify to find the solutions, which are x = 1/2 and x = -3.

Step-by-step explanation:

To determine the solutions to a quadratic equation algebraically, you can use the quadratic formula. The quadratic formula is x = (-b ± √(b^2 - 4ac)) / 2a, where a, b, and c are the coefficients of the quadratic equation in the form ax^2 + bx + c = 0. Let's create a problem and solve it algebraically:

Problem: Solve the quadratic equation 2x^2 + 5x - 3 = 0 algebraically.

Solution:

Step 1: Identify the coefficients a, b, and c:

a = 2, b = 5, c = -3.

Step 2: Substitute the values into the quadratic formula:

x = (-5 ± √(5^2 - 4 * 2 * -3)) / (2 * 2)

Step 3: Simplify the expression inside the square root:

x = (-5 ± √(25 + 24)) / 4

x = (-5 ± √49) / 4

x = (-5 ± 7) / 4

Step 4: Solve for x:

x1 = (-5 + 7) / 4 = 2/4 = 1/2

x2 = (-5 - 7) / 4 = -12/4 = -3

Therefore, the solutions to the quadratic equation 2x^2 + 5x - 3 = 0 are x = 1/2 and x = -3.

Answer 2

Step-by-step explanation:

So a quadratic equation in factored form is generally given in the form:
`a(x-b)(x+c)`. Sometimes x may have a coefficient but I'll just create a simple quadratic equation. In this form it's easy to see that the solutions to the equation are b and -c, because if you input them in as x, you'll get one of the factors equal to 0, and it'll make the y-value zero, thus it's a solution. So b and c can literally be anything. But for this example I'll chose 3 and 6 to keep it simple. Inputting these as b and c you'll get:
`a(x-3)(x+6)`. Now the value of a will determine the stretch/compression, but to keep it really simple I'll just say that a=1. So now all you have to do is expand the two factors. When you expand them you'll get it in the form:
`(x-b)(x+c) = x^2+cx-bx-bc` by using the foil method. So if you expand the two factors in this example you'll get:
`x^2+6x-3x-18` which simplifies to
`x^2+3x-18`. So cool now we have our equation. Now it's time to solve it, you can either factor it (since you literally just had it in factored form), or use the quadratic formula I'll show both methods

Factoring:

When you factor an equation in the form:
`ax^2+bx+c`. You usually multiply the coefficients a and c. Then you'll find factors of ac that add up to b. In this example ac is really just c because a is 1...

So given the equation:
`x^2+3x-18` You want to look for factors of -18 that add up to 3. And whenever you have a negative number as AC you really want to look for factors that have a distance of b which in this case is 3. For example I would say that 9 and 3 have a distance of 6. And then after that depending on whether b is negative or positive. I can choose which factor will be negative and which will be positive. If b was 6 well then 9 would have to be positive and 3 would be negative, but if b was -6 then 9 would have to be negative and 3 would have to be positive.

So in this example look for factors that have "distance" of 3. Well 6 and 3 have a "distance" of 6 so I know those 2 will be the factors. Since 3 is positive that means 6 will be positive, and 3 will be negative. so they add up to 3 and multiply to get -18

AC factors that add up to 3 and multiply to -18: 6 and -3

Now write it in factored form using the two factors:

`(x-3)(x+6)`

Now set up each factor equal to 0 and solve for x:

`x-3 = 0\\x=3`

`x+6=0\\x=-6`

This gives you the two solutions: x=3 and x=-6

Quadratic Formula:

The quadratic formula is usually used when a equation can't be easily factored. The quadratic formula gives both zeroes in the equation:
`x=(-b\pm√(b^2-4ac))/(2a)`. Notice the plus or minus sign next to the square root? That's where the the two solutions will come from, or possibly one solution if it's at the vertex. or maybe even no real solutions if the quadratic never intersects the x-axis.

Ok so let's identify a, b, and c. We're given the equation:
`x^2+3x-18`. so a=1, b=3, and c=-18. Plugging these values into the equation you get:

`x=(-(3)\pm√((3)^2-4(1)(-18)))/(2(1))\\`

Simplifying it a bit gets you the equation:

`x=(-3\pm√(81))/(2)`

Simplifying the radical gives you:

`x=(-3\pm9)/(2)`

Use the positive sign:

`x=(-3+9)/(2)\\x=(6)/(2)\\x=3`

Use the negative sign:

`x=(-3-9)/(2)\\x=(-12)/(2)\\x=-6`

This gives you the two solutions x=3, and x=-6