How do you factor x squared + 2x + 5? Thanks!
2006-12-07 10:01:01 · 5 answers · asked by teef 1 in Science & Mathematics ➔ Mathematics
To factorise a quadratic equation ax^2 + bx + c where you cannot use cross multiplication, use the formula
x = [-b ± √(b^2-4ac)]/2a
In this case, you have a problem on factorising x^2 + 2x + 5. So here's the solution:
x = [-2 ± √(2^2-4*1*5)]/2*1
Notice the discriminant in the square root term (2^2 - 4*1*5)? When you calculate 2^2 - 4*1*5, you will get -16. Any negative value in the square root term gives an imaginary root (or in the simplest definition, you get no real roots, that is no answer).
If you try pressing √-16, you'll get an error.
Therefore, I conclude that this equation has imaginary roots, or no real roots. Just write your answer as no real roots.
This is a real fact, and it doesn't mean that I'm not giving any answers or I'm unable to come up any answers. The real fact is that this equation cannot be factorised.
2006-12-07 20:25:20 · answer #1 · answered by horensen 4 · 0⤊ 0⤋
eh, its not really factorable as a whole, but you CAN do.... (x(x+2))+5 where as i've taken 1x from both x^2 and 2x... leaving x+2 inside parenthesis, and outside would still have the 5, but now we've added x from our factoring good luck :) edit: i checked my work and changed it, sorry jumped to conclusions a little too fast. If you need any more help of you dont understand what i did, OR you think i did it wrong, please edit, and i'll fix it accordingly heavenlyangel1... grow up, you can't get an answer like that and you know it, enjoy the report, and thanks for the thumbs down.
2016-05-23 04:40:32 · answer #2 · answered by ? 4 · 0⤊ 0⤋
x^2 + 2x + 5
Factors would be of the form (x + a)(x + b) and the only choice for a & b is 1 & 5 and you can't get the coefficient of the x term with them
Solving the equation
x^2 + 2x + 5
using the quadratic formula yields
x = - 1 +- 2i
2006-12-07 10:12:51 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Use the quadratic formula ax^2+bx+c = 0 has solutions x = (-b +/- sqrt(b^2-4ac))/2a. In you case, if x^2+2x+5 = 0, then x = (-2 +/- sqrt(4-20))/2 = -1 +/- sqrt(-16)/2 = -1 +/- 2i, where i = sqrt(-1). It is complex. Therefore x^2+2x+5 = (x+1+2i)(x+1-2i).
2006-12-07 10:11:46 · answer #4 · answered by Anonymous · 0⤊ 0⤋
set the problem equal to zero and use the quadratic formula
x= -1 +_ 4i
2006-12-07 12:47:33 · answer #5 · answered by monica 1 · 0⤊ 0⤋