x² + 10x + 13 = 0
x² + 10x + 25 = -13 + 25
So (x + 5)² = 12
So x + 5 = ±√12
(Note first of all the number that there are TWO numbers that when squared give 12 and they are √12 and - √12 Hence the ±√12)
So x = -5 ± 2√3 (Note that √12 = √(4 * 3) = √4 *√3 = 2√3)
Luis U - Delta = (10)² - 4 * (1)* 13 ( a = 1 not 2)
= 100 - 52
= 48
So roots are ½(-10 ± √48)
ie ½(-10 ± √(16 * 3))
ie ½(-10 ± 4√3)
ie -5 ± 2√3
2007-01-09 12:23:43
·
answer #1
·
answered by Wal C 6
·
0⤊
0⤋
a million. be sure your lead coefficient is the same as one. even if it isn't, be sure you aspect out in spite of the lead coefficient is. x^2 + 10x + 13 = 0 examine! 2. Slap parenthesis around the x values (x^2 + 10x) + 13 = 0 examine! 3. Take the coefficient of x, halve it, then sq. it. (10/2)^2 (5)^2 25 examine! 4. Plug it in to the equation, yet be sure you subtract the same volume from the different fringe of the equation. (x^2 + 10x + 25) + 13 - 25 = 0 stop precise there! once you've a range outdoors the parenthesis, be sure you multiply the range with information from that variety, THEN subtract it. e.g. 2(x^2 + 4x + 4) + 3 - (2 x 4) not 2(x^2 + 4x + 4) + 3 - 4 examine! 5. Simplify. (x^2 + 10x + 25) + 13 - 25 = 0 (x^2 + 10x + 25) - 12 = 0 (x + 5)^2 - 12 = 0 in case you've been to graph it, you would possibly want to shift the vertex left 5 areas, down 12 areas, and then graph a parabola pointing up and conventional sized. desire that helped!
2016-12-02 01:44:31
·
answer #2
·
answered by ? 4
·
0⤊
0⤋
x^2 + 10x + 13 = 0
x^2 + 10x = -13
x^2 + 10x + 25 = -13 +25
(x+5)^2 = 12
x+5 = 12^(1/2)
x = 12^(1/2) - 5
2007-01-09 12:20:39
·
answer #3
·
answered by Bigsky_52 6
·
0⤊
0⤋
1first pass the constant (in this case 13) to the other side: x^2 = 10x= -13
2 Get K or (b/2)^2 (in this case 10) so it is 25
Add k (25) to both sides: x^2 + 10x + 25 = 12
now simplify (factor): (x + 5)^2 = 12
get sq root (both sides) : x + 5 = +/- (sq root 12)
isolate x: x = -5 +/- (sq root 12)
hope it helps
2007-01-09 12:31:01
·
answer #4
·
answered by mazp66 3
·
0⤊
0⤋
x^2 + 10x + 13 = 0;
delta = 10*10 - 4*2*13;
delta = 100 - 104;
delta = -4;
x = (-10- 2i)/4 or x = (-10+2i)/4;
x^2 + 10x + 13 = 2(x-(-10- 2i)/4)(x-(-10+2i)/4);
2007-01-09 12:22:58
·
answer #5
·
answered by Luis U 2
·
0⤊
0⤋
> - 1.5359
> - 8.464
(-1.5359)^2 + (10 * -1.5359) +13 = 0
2.359 - 15.359 + 13 = 0
(-8.464)^2 + (10 * 8.464) + 13 = 0
71.64 - 84.64 + 13 = 0
2007-01-09 14:22:34
·
answer #6
·
answered by robertonereo 4
·
0⤊
0⤋
x= -23
2007-01-09 12:20:37
·
answer #7
·
answered by Crystal 1
·
0⤊
0⤋
(x+5)^2=12
x=-5+sqrt(12)
or -5-sqrt(12)
2007-01-09 12:19:48
·
answer #8
·
answered by well thts it...... 3
·
0⤊
0⤋