tell me how to factorise quadratic equation using formula?

Question

the formula is b+ or - (b^2-4ac)^.5 / 2a

in which

a^2-68a+256=0
a=1 ; b = -68 ; c = 256

Answer 1

Your formula is not completely well
If ax^2 + bx + c = 0 then
x1 = {b - root(b^2-4ac)}/(2a) and
x2 = {b + root(b^2-4ac)}/(2a).
When you know this then
ax2 + bx + c = a(x - x1)(x - x2) and this what you want.
If b^2-4ac < 0 then factorizing is impossible for reel numbers.

Answer 2

b+ or - (b^2-4ac)^.5 / 2a <= this formula is used to solve the equation.

if u want to solve it for x the ans is

x = [ (-68) + or - ( 68^2 - 4(1)(256) ) ^ 0.5 ] / 2

u will get x1= 4 and x2= 64

Then

a^2-68a+256 = (a - x1)(a - x2) =0

so

a^2-68a+256 = (a - 4)(a - 64)

right?

Answer 3

This also means solving for x in the quadratic equation
ax² + bx + c = 0
where a,b and c are real numbers and a ≠ 0

Very well...
ax² + bx + c = 0

Transpose c
ax² + bx = -c

Divide the equation with a
x² + b/a x = -c/a

add b²/4a² to both sides
x² + b/a x + b²/4a² = -c/a + b²/4a²

Simplify the right side
x² + b/a x + b²/4a² = (-4ac + b²)/4a²

factor the left side (perfect square trinomial)
(x + b/2a)² = (b² - 4ac)/4a²

Get sqrt of both sides
x + b/2a = ±√(b² - 4ac)/√(4a²)

Get sqrt of denominator (right side)
x + b/2a = ±√(b² - 4ac)/2a

Transpose b/2a
x = -b/2a ±√(b² - 4ac)/2a

combine into 1 expression
x = [-b ± √(b² - 4ac)]/2a
where a ≠ 0

the method of getting the "quadratic equation" is called "completing the square." You can also use this method in any other quadratic equations.

^_^

Answer 4

X^2-3x=10 Subtract 10 from each and each area X^2-3x-10 Now you've your a,b, and c for the formula. A=a million B=-3 C=-10 The Quadratic formula is: -b plus or minus the sq. root of [b^2-(4ac)] and then all of that over 2a. you should have 2 solutions because you subtract the sq. root quantity in a unmarried, and upload it the subsequent time you workout consultation the region.

Answer 5

a^2 - 68a + 256 = 0

a = (-b ± sqrt(b^2 - 4ac))/2a

a = (-(-68) ± sqrt((-68)^2 - 4(1)(256)))/(2(1))
a = (68 ± sqrt(4624 - 1024))/2
a = (68 ± sqrt(3600))/2
a = (68 ± 60)/2
a = (8/2) or (128/2)
a = 4 or 64

ANS : (a - 4)(a - 64)

Answer 6

plug the numbers in for a, b and c and then solve.