solve the equation xsquared+x-12=0?

x^2 + x - 12 = 0

Do the quadratic equation.

a = 1

b = 1

c = -12

x+y=4, x^2 + y^2 = 40 Take the first equation and isolate y: y=4-x Since that IS y, plug it into the second equation for y: x^2 + (4-x)^2 = 40 x^2 + 4^2 -8x + x^2 = 40 x^2 + 16 - 8x + x^2 = 40 2x^2 - 8x + 16 = 40 2 (x^2 - 4x + 8) = 40 x^2 - 4x + 8 = 20 x^2 - 4x - 12 = 0 (x-6)(x+2) = 0 So x is 6 or -2. (You could use the quadratic formula for that, too.) We have two answers for x, so plug both in for x in the original equation: 1) 6 + y = 4; y = -2 2) -2 + y = 4, y = 6 Do they make sense? 6^2 + -2^2 = 36 + 4 = 40 -6^2 + 2^2 = 36 + 4 = 40 Yup! Wolfram Alpha can give you the answer but not show you the steps in this particular case, but it's just isolating one variable and plugging it into the other equation. When you do that, though, just be careful that you don't DIVIDE both sides by an expression that could equal zero. If you do that, you'll wander into undefined results, which is where you get those weird "1=2" proofs you see on the net every now and then.

3

12

X^2 + X -12 = 0 is a quadratic equation of the form "AX^2 +BX + C = 0" where the terms are A=1, B=1 and C=-12. The equation has two solutions which are given by the formula:

(B +/- SQRT(B^2 - 4A*C))/2*A

Substitution yields:

(1 + SQRT(1 + (-4)*(-12))/2*1

(1 - SQRT(1 + (-4)*(-12))/2*1

X = (1 +/- SQRT(25))/2

x = (1 + 5)/2 and (1 - 5)/2 or 3 and -4

The equation is therefore (x + 3) (x - 4) = 0

By Factoring

( x+4 )( x-3 )-12=0

therefore: X= -4, X= 3

Checking: when x= -4

( -4)(-4) + (-4)-12=0

16-16=0

0=0

when X= -3

(3)(3) + (3) -12=0

9+3-12=0

0=0

x squared plus x = 12, so x =3! 3x3+3-12=0! x equals 3!

(x+4)(x-3)

or x=-4

x^2+x-12 =0

just factor it out.

(x-3) (x+4)..therefore x=3 or x= -4

(x+4)(x-3)=0

x+4=0

x= -4

x-3=0

x=3

SS(solution set/ set builder notation):(3,-4)

you can check the answer by substituting 3 and -4 with x in the original equation