Does anyone Have a Clue How to Do these?

Question

Write a quadratic equation in standard form and then solve.

4.)y=2x^2-2x-12

6.)y=x^2+x-4

Use the quadratic formula to solve.

9.)x^2+12x+36=0


Please Show work I'M realy stumped on how to do these!!

Answer 1

Write a quadratic equation in standard form and then solve.

4.)y=2x^2-2x-12
0=2x^2-2x-12
x^2-x-6=0
(x-3)(x+2)=0
x-3=0
x=3
x+2=0
x=-2

6.)y=x^2+x-4
x^2+x-4=0
x=(-1+/-√(1+64))/2
x=-1/2+1/2 √65
x=-1/2-1/2 √65

Use the quadratic formula to solve.

9.)x^2+12x+36=0
x=(-12+/-√(144-36*4))/2
x=(-12+/-√0)/2
x=-6

Answer 2

ok, the quadratic formula is -b (+ or -) square root of (b^2-4ac) divided by 2a
alrighty, for # 4 a=2 b=-2 and c=-12
type that into the calculator, first adding then subtracting to find what x = when y is 0. that is the answer. Yay! (by the way im 14 and i get this perfectly. and also I have the formula memorized)

Answer 3

People have shown you how to solve the first two equations, provided you make y=0.

What has been missed is that the question is flawed. Each equation IS already a quadratic equation. Each equation is already in standard form (unless your teacher has some strange idea that to say "standard form" means make y=0!).

Each equation is in two variables - each cannot be "solved"! For example the first equation allows any x value from negative infinity to positive infinity, and any y value from -12.5 to positive infinity - it makes no sense to talk of a "solution" to this equation.

As each equation describes a parabola if plotted, you can find certain significant points on such a parabola but that is not "solving" the equation itself.

Answer 4

If you know the quadratic formula, it's simple. If you don't know the formula, showing the work would be useless.

Answer 5

4. [2 +-sqrt(4+96)]/4 or x=-2 or x=3

6. [-1 +-sqrt(1+16)]/2

7. [-12 +-sqrt(144-144)]/2 or x=-6 double root