quadratic eqn: no solution because negative?

[sarahj3]

I am pretty sure these would not have an answer because they come out negative

3x^2=9x-7

if you plug into the quadratic formula it equals

x1x2=9+or-the square root of -165/6

can't do because negative?

also

25(x-3)^2=100

divide by 25=
(x-3)^2=4
x^2-9=4
x^2-9-4=0
pluged into quadratic formula
=9+or-the square root of -65/2

 

[skeeter]

\(\displaystyle 3x^2 = 9x - 7\)
\(\displaystyle 3x^2 - 9x + 7 = 0\)

a = 3, b = -9, c = 7

\(\displaystyle b^2 - 4ac = 81 - 4*3*7 = -3\) ... since the discriminant < 0, the equation has two imaginary roots

\(\displaystyle \L x = \frac{9 \pm i \sqrt{3}}{6}\)


\(\displaystyle 25(x - 3)^2 = 100\)
\(\displaystyle (x - 3)^2 = 4\)
\(\displaystyle x - 3 = \pm 2\)
\(\displaystyle x = 3 \pm 2\)
x = 5 or x = 1

 

[Denis]

Sarah, you need to get back to basics...

"3x^2=9x-7
if you plug into the quadratic formula it equals
x1x2=9+or-the square root of -165/6"

(-9)^2 = 81, not -81
AND what's "x1x2"? Should be: x = [9 +or- sqrt(-3)] / 6
YES, the brackets are required.

"25(x-3)^2=100
divide by 25=
(x-3)^2=4
x^2-9=4
x^2-9-4=0
pluged into quadratic formula
=9+or-the square root of -65/2"

(x - 3)^2 = x^2 - 6x + 9 : NOT x^2 - 9
SO: x^2 - 6x + 9 = 4
x^2 - 6x + 5 = 0