One possibility is 7/2 + sqrt(1/4)
Without an equality sign the given expression can't be considered to be an equation and so therefore there are no solutions.
You did not state the operation: add or subtract. Is it 2x + 3 + 4 = 17? Is it 2x - 3 + 4 = 17?
If you mean: y = 3x-4 then it is a straight line equation whose slope is 3 and its y intercept is -4
I gotchu homie: It's The equation has x = 4 and x = -4 as its only solutions.
None because without an equal it is not an equation. But if it was in the form of x2+7x+12 = 0 then it would have 2 solutions which are x = -3 and x = -4
It is a quadratic equation and can be rearranged in the form of:- x2-x-6 = 0 (x+2)(x-3) = 0 Solutions: x = -2 and x = 3
4 radical 3 is about 6.92820323
Without an equality sign the given expression can't be considered to be an equation and so therefore there are no solutions.
They are actually to the one half power. You can take a factor in the radical and sqrt it and put in on the outside... Ex. sqrt(28) = sqrt(4 * 7) = sqrt(22 * 7) = 2sqrt(7) sqrt(28) = 2 * sqrt(7)
To find an equation where two numbers are solutions of x, do the following: x=-3 , x=4 x+3=0 , x-4=0 Multiply them together and set equal to zero, i.e. (x+3)(x-4) = x2-x-12 = 0
If the discriminant of a quadratic equation is less then 0 then it will have no real solutions.
radical(48)/radical(3) = radical(48/3) = radical(16) = 4 Technically, radical(16) is +4 OR -4 but in such questions often only the principal root is required.
You did not state the operation: add or subtract. Is it 2x + 3 + 4 = 17? Is it 2x - 3 + 4 = 17?
-3
First, note that radical 4 is 2. So 3xradical 4 is just 6, Now we have 6+2 radical 3. You can't do much with this except factor out a 2 if you want 2(3+Radical 3)
If you mean: y = 3x-4 then it is a straight line equation whose slope is 3 and its y intercept is -4
I gotchu homie: It's The equation has x = 4 and x = -4 as its only solutions.