The book asks to us the quadratic formula to solve for the equation.
Can someone help with these examples:
1.) x^2+4x-2=0
2.) 2x^2 -5x-2=0
3.) x^2+2x=4x
4.) -6x^2 +3x+2=3
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There are more problems but I think I can figure it out once someone helps me with the above examples. Gracias
Can someone help me show their work so I can see visual steps to work out these kinds of questions.
think back to algebra I. The QF is
x = [-b +/- sqrt(b^2-4a)]/2a
So, just plug in your number. The first is
x = [-4 +/- sqrt(16-4(1)(-2)]/2
= [-4 +/- sqrt(16+8)]/2
= -4 +/- sqrt(24)]/2
= -2 +/- sqrt(6)
Do the same with the others. First step is to get everything on the left side so you can see what a,b,c are.
Sure! I can help you solve those equations using the quadratic formula. The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / 2a
Let's solve each equation step by step:
1.) x^2+4x-2=0
In this equation, a = 1, b = 4, and c = -2. Substitute these values into the quadratic formula:
x = (-4 ± √(4^2 - 4(1)(-2))) / 2(1)
Simplify:
x = (-4 ± √(16 + 8)) / 2
x = (-4 ± √24) / 2
Now, simplify the square root:
x = (-4 ± √(4 * 6)) / 2
x = (-4 ± 2√6) / 2
Simplify further:
x = -2 ± √6
Hence, the solutions are x = -2 + √6 and x = -2 - √6.
2.) 2x^2 -5x-2=0
Here, a = 2, b = -5, and c = -2. Substitute these values into the quadratic formula:
x = (-(-5) ± √((-5)^2 - 4(2)(-2))) / 2(2)
Simplify:
x = (5 ± √(25 + 16)) / 4
x = (5 ± √41) / 4
Hence, the solutions are x = (5 + √41) / 4 and x = (5 - √41) / 4.
3.) x^2+2x=4x
Rearrange the equation to bring all terms to one side:
x^2 + 2x - 4x = 0
x^2 - 2x = 0
Now, factor out x from both terms:
x(x - 2) = 0
This equation will be true if either x = 0 or x - 2 = 0. Solve for x in each case:
x = 0 or x - 2 = 0
If x - 2 = 0, then x = 2.
Hence, the solutions are x = 0 and x = 2.
4.) -6x^2 +3x+2=3
Here, a = -6, b = 3, and c = -1. Substitute these values into the quadratic formula:
x = (-3 ± √((3)^2 - 4(-6)(-1))) / 2(-6)
Simplify:
x = (-3 ± √(9 - 24)) / -12
x = (-3 ± √(-15)) / -12
Since the discriminant (the term inside the square root) is negative, the equation has no real solutions.
I hope this helps you solve your quadratic equations. Let me know if you have any further questions!
Certainly! I'd be happy to help you solve these quadratic equations using the quadratic formula. The quadratic formula is a mathematical formula used to find the solutions of a quadratic equation of the form ax^2 + bx + c = 0. The formula is:
x = (-b ± √(b^2 - 4ac)) / (2a)
Let's solve the examples one by one:
1.) x^2 + 4x - 2 = 0
Here, a = 1, b = 4, and c = -2.
Plug these values into the quadratic formula:
x = (-4 ± √(4^2 - 4(1)(-2))) / (2(1))
Simplify and evaluate the expression within the square root:
x = (-4 ± √(16 + 8)) / 2
x = (-4 ± √24) / 2
x = (-4 ± 2√6) / 2
Simplify further:
x = -2 ± √6
So, the solutions are: x = -2 + √6 and x = -2 - √6
2.) 2x^2 - 5x - 2 = 0
Here, a = 2, b = -5, and c = -2.
Applying the quadratic formula:
x = (-(-5) ± √((-5)^2 - 4(2)(-2))) / (2(2))
Simplify:
x = (5 ± √(25 + 16)) / 4
x = (5 ± √41) / 4
No further simplification can be done, so the solutions are:
x = (5 + √41) / 4 and x = (5 - √41) / 4
3.) x^2 + 2x = 4x
First, move all the terms to one side of the equation, so it becomes:
x^2 + 2x - 4x = 0
x^2 - 2x = 0
Now, factor out an x:
x(x - 2) = 0
This equation is already factored, so the solutions are:
x = 0 and x = 2
4.) -6x^2 + 3x + 2 = 3
First, rearrange the equation to bring it to the standard quadratic form:
-6x^2 + 3x - 1 = 0
Here, a = -6, b = 3, and c = -1.
Apply the quadratic formula:
x = (-3 ± √(3^2 - 4(-6)(-1))) / (2(-6))
Simplify:
x = (-3 ± √(9 - 24)) / -12
x = (-3 ± √(-15)) / -12
Since there is a negative value under the square root, we have complex solutions.
x = (-3 ± i√15) / -12
Simplify further:
x = (1 ± i√15) / 4
So, the solutions are: x = (1 + i√15) / 4 and x = (1 - i√15) / 4