solve X^2 + 12 X + 32 = 0?

Question

solve X^2 + 12 X + 32 = 0

Answer 1

X^2 + 12 X + 32 = 0 first factor

(x+8)(x+4) set each eqaul to 0

x+8=0

x= -8

x+4=0

x=-4

Answer 2

X = - 4 and - 8.

Here's how these solutions are found. The original quadratic expression factorises:

X^2 + 12 X + 32 = (X + 8) (X + 4). So

X^2 + 12 X + 32 = 0 ==> (X + 8) (X + 4) = 0.

So one or other of these parenthetical expressions must be zero.

Therefore X = - 8 or - 4.

CHECK: For X = - 8, X^2 + 12 X + 32 = 64 - 96 + 32 = 0;
and for X = - 4, X^2 + 12 X + 32 = 16 - 48 + 32 = 0.

So both of these values are indeed solutions.

Live long and prosper.

Answer 3

In the general case, this can be solved with the quadratic formula. ttp://mathworld.wolfram.com/QuadraticEquation.html
But, this particular one is easily solved without it:

Think about the roots of 32 - You need roots of 32 which will sum to the middle term (in this case 12) Well, 32 is 2*16, and also 4*8...

(x+4)*(x+8) = x^2 + 12x + 32 = 0

so, if that's the case, either (x+4) = 0, or (x+8) = 0, so x is either -4 or -8.

Answer 4

X^2 + 12 X + 32 = 0

look for two number that will multiply to give 32 and check if the two numbers add up to 12

(x + 4)(x + 8) = 0

x: -4 or -8

there is your answer

Answer 5

solve X^2 + 12 X + 32 = 0
or(x+8(x+4)=0
answers: -8.-4.

Answer 6

x^2+12x+32=0
=> x^2+4x+8x+32=0
=>x(x+4)+8(x+4)=0
=>(x+4)(x+8)=0
therefore either x+4=0 or x+8=0
Hence x= -4 or -8

Answer 7

(x+4)(x+8) = 0
(x+4) = 0
x = -4
(x+8) = 0
x = -8
Thus, from factoring the quadratic equation, x is equal to -8 and -4.

Answer 8

factor

look for two numbers that multiply to 32 and add to 12

8 and 4

(x+8) (x+4) = 0

x= -8 or -4

Answer 9

Factor the equation to get
(x+8)(x+4)=0.
Now set each factor to 0 and solve.
The solutions are
x = -8 and x = -4.

Answer 10

Problem : ( log x )² + log x^(-1) - 12 = 0 Let log x = y => y² - y - 12 = 0 => (y - 4)(y + 3) = 0 => y = 4 ; y = -3 => log x = 4 ; log x = -3 => x = 10^4 ; x = 10^(-3) => x = 10000 ; x = 0.001 => (Ans)