Help using discriminant to determine number and nature of the solutions to quadratic: x^2 - 12x - 18 = 0?

Answer 1

x = [ 12 ±√(144 + 72) ] / 2
x = [ 12 ± √(216) ] / 2
x = [12 ± 2√(79) ] / 2
x = [ 6 ± √(79) ] / 2 gives two solutions.

Answer 2

The quadratic formula is:
x={-b+/- sq.rt.(b^2-4ac)} /2a
b^2-4ac is the Discriminant. Not sq.rt.(b^2-4ac),just
the b^2-4ac term.
If that term is positive, then you will end up with 2 real and unequal roots. That's because you can take the square root of a positive number, getting a + and a -solution. These 2 solutions then combine with the other terms in the formula to yield the answers.
If b^2-4ac=0, then the solution is x=-b/2a. Since a
quadratic always has 2 solutions, we say there are
2 real,equal roots.
If b^2-4ac is negative, there is no solution possible
for the real number system, because you cannot
take the square root of a negative number.

For x^2-12x-18,
a=+1, b= -12, c= -18
b^2-4ac=(-12)^2-(4)(1)(-18)
b^2-4ac=144+72, or 216
b^2-4ac is positive, so 2 unequal roots exist in the
Real Number System

Answer 3

Given: (x^2)-(12x)-18=0
where a=1, b= -12 & c= -18

following the formula of discriminant,
D=(b^2)-(4ac)
D=(-12^2)-(4*1*-18)
D=144-(-72)
D=144+72
D=216

Therefore the discriminant has 2 real and distinct solutions.

since that when D>0, their is 2 real & distinct solutions
when D=0, their is 1 repeating & real Solution
when D<0, their is no real solution

Answer 4

The discriminate is b²-4ac which is 12²-4(1)(-18) = 216 since the discriminate is a positive value the quadratic equation will have two real roots.

Answer 5

216>0

Answer 6

D = b^2 - 4ac

D = (-12)^2 - 4(1)(-18)

D = 144 + 72 = 216 > 0

So there are two distinct real solutions.

Answer 7

Using quadratice formula, x=(12+/-sqr(12^2-(4*1*-18))) /(2*1).
Since (12^2-(4*1*-18))>0, it contains 2 real solutions, if thats what your asking for.

Answer 8

The discriminant is:
b^2 - 4ac
= 144 - (4*1*(-18))
= 216 > 0

Since the discriminant is greater than zero (i.e. it's positive), that means that there are two distinct real roots.

Answer 9

x = 6 - square of 54 or x=6 + square of 54 (for example square of 36 is 6 or square of 25 is 5 ) from shayan from Iran

Answer 10

6+or-[(inv)56^2]