quadratic equation, roots

Solutions of quadratic equation are:

x = ((-b) +- (sqrt(b^2 - 4ac)))/2a

One of the solution is 1.5 and a=1, b=m and c=24

So

1.5 = (-m + (sqrt(m^2 - 4(1)(24)))) / 2(1)
3 = -m + sqrt(m^2 - 96)
3 + m = sqrt(m^2 - 96)
(3+m)^2 = m^2 - 96
9+m^2+6m = m^2 - 96
9+6m = - 96
6m = -96 -9
m = - (105/6)
m = - 17.5

Is the answer -17.5?

yes but - 17.5
THKS

opps.. made a tiny little mistake..

I have edited my previous post.

Since x=1.5 is on the roots substitute x in the equation,

(1.5)^2+1.5m+24=0

26.25+1.5m=0

m=-26.25/1.5

m=-17.5

yes. Its -17.5. I did it in this way:

1.5^2+1.5x+24=0, Or
1.5(1.5+x)=-24
1.5+x=-24/1.5=-16
x=-16-1.5=-17.5

all the same, but avoid the complicacy of squiring the decimal 1.5.

thanks.

yes. Its -17.5. I did it in this way:

1.5^2+1.5x+24=0, Or
1.5(1.5+x)=-24
1.5+x=-24/1.5=-16
x=-16-1.5=-17.5

all the same, but avoid the complicacy of squiring the decimal 1.5.

thanks.

francoisph wrote:If one of the roots of the quadratic equation x2 + mx + 24 = 0 is 1.5, then what is the value of m?

I did it a different way.

If one root is 1.5, then we can write the quadratic as:

(x - 1.5)(x + y) = 0

From the equation give, we know that:

(-1.5)(y) = 24

So:

y = 24/(-3/2) = 24 * (-2/3) = -16

So:

(x - 1.5)(x - 16) = x^2 + mx + 24

m simply equals (-1.5) + (-16) = -17.5

francoisph wrote:If one of the roots of the quadratic equation x2 + mx + 24 = 0 is 1.5, then what is the value of m?

Sum of the roots = -m -- eq 1

product of roots= 24

Let two roots be a, b: (1.5,b)

1.5+b = -m

1.5 * b = 24

b=16

so applying eq 1:

1.5+ 16 = -m

m = -17.5

sub x=1.5 .in x2 + mx + 24 = 0

(1.5)^2+1.5m+24=0
2.25+24+1.5m=0
1.5m=-26.25
m=-17.5

francoisph wrote:If one of the roots of the quadratic equation x2 + mx + 24 = 0 is 1.5, then what is the value of m?