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
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?
9 answers
so what is the 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)
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)
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) )
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 :)
(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.
You get them in your first problem and in your third problem.
3 answers