How Do You Use Quadratic Equation To Solve X^2-5x-24=0?

x = [ - b ± √ ( b ² - 4 a c ) ] / 2 a

x = [ 5 ± √ ( 25 + 96 ) ] / 2

x = [ 5 ± √ ( 121 ) ] / 2

x = [ 5 ± 11 ] / 2

x = 8 , x = - 3

x² - 5x - 24 = 0
x² - 5/2x = 24 + (- 5/2)²
x² - 5/2x = 96/4 + 25/4
(x - 5/2)² = 121/4
x - 5/2 = ± 11/2

x = 11/2 + 5/2, x = 16/2, x = 8
x = - 11/2 + 5/2, x = - 6/2, x = - 3

Answer: x = 8, - 3; factors: (x - 8)(x + 3)

a million) 4x^2+8x-y=-4 3x^2-2x-y=5 Subtracting the above equations, 4x^2-3x^2+8x+2x=-9 x^2+10x+9=0 x^2+9x+x+9=0 x(x+9)+a million(x+9)=0 (x+9)(x+a million)=0 for this reason, x=-9 or -a million Substituting those values contained in the first equation, even as x=-9, 4(-9)^2+8(-9)+4-y=0 4(80 one)+(-seventy 2)+4-y=0 324-seventy 2+4-y=0 -y=-256 for this reason, y=256 even as x=-9 even as x=-a million, 4(-a million)^2+8(-a million)-y=-4 4-8+4-y=0 -y=0 for this reason even as x=-a million, y=0 for this reason, the options to those equations, even as x=-9, y=256 even as x=-a million, y=0 2)x^2-y=-a million 2x^2+3y=28 Multiply the first equation by technique of three, 3x^2-3y=-3 upload the above equation with the 2d equation, 2x^3+3x^2=28-3 5x^2=25 X^2=5 for this reason, x=+squareroot5 or -squareroot5 even as x=+ or - squareroot 5, Equation a million turns into, 5-y=-a million -y=-6 for this reason, y=6 for this reason, the answer to those equations, X=+ or - squareroot5, y=6 3)4x^2-2y^2=4 X^2+y^2=10 Multiply the 2d equation by technique of two, 2x^2+2y^2=20 upload the above and the first equations, 4x^2+2x^2=4+20 6x^2=24 X^2= 4 for this reason, x=+ or-2 even as x=+ or-2, Equation 2 turns into, 4+y^2=10 Y^2=6 for this reason, y= + or-squareroot 6 for this reason, the answer to those equations, X=+ or -2, y =+ or - squareroot6

x^2 - 5x - 24 = 0
x^2 + 3x - 8x - 24 = 0
(x^2 + 3x) - (8x + 24) = 0
x(x + 3) - 8(x + 3) = 0
(x + 3)(x - 8) = 0

x + 3 = 0
x = -3

x - 8 = 0
x = 8

∴ x = -3, 8

If the given equation is in the form of ax^2+bx+c=0 then

the roots of the equation are

x= (-b±√(b^2-4ac))/2a

Here a=1;b=-5;c=-24;

x=8, -3

X^2-5x-24=0

for equation AX^2+BX+C = 0
http://upload.wikimedia.org/math/3/e/a/3ea647783b5121989cd87ca3bb558916.png

so here, a = 1
as b = -5
c = -24

plug it into the equation and you will get X :)

Use the quadratic equation:

((-b +/- root( b^2-4ac) ) / 2a

Where a = 1, b = - 5 c= -24

These numbers are the coefficients of each term.


I think you should get x = 8 and x=-3

Take careful note of where the brackets are in the formula, and of minus numbers. The plus or minus at the start gives you two solutions.