Solve each equation and leave the answer in simplest radical form.
x squared+6x-9=0
n squared-2n+2=5
3y squared-4y-2=0
I never learned how to do this. Can you help me?
I am going to do the second one. Check the first one carefully for typos
n^2 - 2 n - 3 = 0 subtracting 5 from both sides
(n-3)(n+1) = 0 factoring
then if n-3 = 0 the equation is satisfied
so
n = 3 adding 3 to both sides
also if n+1 = 0, the equation is satisfied
so
n = -1 subtracting one from both sides
Normally you will find two solutions for a quadratic. In this case they are 3 and -1
so what is the radical form?
There is no radical in that solution for n because the solutions are whole numbers.
for the first question, can the answer be 9 and 7?
The third one will have a radical solution as well as the first
3 y^2 - 4y -2 = 0
3 y^2 - 4y = 2 adding 2 to both sides
y^2 - (4/3) y = 2/3 dividing by coef of x
y^2 - (4/3)y + 4/9 = 2/3 + 4/9 = 10/9 adding the square of half the coef of x to both sides to make a perfect square
(y-2/3)^2 = 10/9
so
y - 2/3 = sqrt (10/9) = (1/3)sqrt (10)
or
y-2/3 = -sqrt(10/9) = -(1/3) sqrt(10)
so
y = (1/3)( 2+sqrt(10) )
or
y = (1/3)( 2-sqrt(10) )
so which one is the radical form?
To find out if 7 or 9 will work in the first one, put 7 for example back in the original equation
(7)^2 +6(7) -9 = 0 ?
No way! So 7 is not a solution for x and you will have to do that whole completing the square thing I did or use the quadratic equation and the solution will have radicals that have to be simplified as I did. (This is why I asked if it is a typo but seeing as how you are doing radicals anyway it was probably not a typo :)
I use the symbol "sqrt" to mean square root or "radical" with index 2
You get them in your first problem and in your third problem.
Of course! I'd be happy to help you solve these equations.
To solve quadratic equations like the ones you provided, we can use the quadratic formula:
x = (-b ± √(b² - 4ac)) / (2a)
In general, quadratic equations can be written in the form ax² + bx + c = 0, where a, b, and c are constants. Now let's solve each equation step by step.
1. x² + 6x - 9 = 0
In this equation, a = 1, b = 6, and c = -9. Plugging these values into the quadratic formula:
x = (-6 ± √(6² - 4(1)(-9))) / (2(1))
x = (-6 ± √(36 + 36)) / 2
x = (-6 ± √72) / 2
Simplifying further:
x = (-6 ± 6√2) / 2
x = -3 ± 3√2
So the solutions to this equation are x = -3 + 3√2 and x = -3 - 3√2.
2. n² - 2n + 2 = 5
Rearranging the equation to match the standard quadratic form, we have n² - 2n - 3 = 0. Now, applying the quadratic formula:
n = (-(-2) ± √((-2)² - 4(1)(-3))) / (2(1))
n = (2 ± √(4 + 12)) / 2
n = (2 ± √16) / 2
Simplifying further:
n = (2 ± 4) / 2
n = 3 or n = -1
So the solutions to this equation are n = 3 and n = -1.
3. 3y² - 4y - 2 = 0
Here, a = 3, b = -4, and c = -2. Plugging these values into the quadratic formula:
y = (-(-4) ± √((-4)² - 4(3)(-2))) / (2(3))
y = (4 ± √(16 + 24)) / 6
y = (4 ± √40) / 6
Simplifying further:
y = (4 ± 2√10) / 6
y = (2 ± √10) / 3
So the solutions to this equation are y = (2 + √10) / 3 and y = (2 - √10) / 3.
Each of these equations has been solved using the quadratic formula, and the solutions are expressed in simplest radical form.
Now if that first one is not a typo, solve it by either completing the square or with the quadratic equation.
I will do completing the square
x^2 + 6 x = 9 adding 9 to both sides to get only 1x^2 and a coefficient of x on the left
then take half of 6 and square it and add to both sides
x^2 + 6x + 9 = 18 made perfect square by adding the square of half the coef of x to both sides
then
(x+3)^2 = 18 which is 9*2
so
x+3 = sqrt (18) = 3 sqrt (2)
or
x+3 = -sqrt(18) = -3 sqrt (2)
so
x = -3+3sqrt(2) = 3(sqrt(2)-1)
or
x = -3 -3sqrt(2) = -3 (sqrt(2)+1)