Question

Which equation has the solutions 2x ^2 +6x+9=0

Answer 1

For this case we have a quadratic equation of the form:

$ax ^ 2 + bx + c = 0$

Where:

$a = 2\\b = 6\\c = 9$

Its roots are given by:

$x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}\\x = \frac {-6 \pm \sqrt {(6) ^ 2-4 (2) (9)}} {2 (2)}\\x = \frac {-6 \pm \sqrt {36-72}} {4}\\x = \frac {-6 \pm \sqrt {-36}} {4}$

By definition we know that:

$i = \sqrt {-1}\\i ^ 2 = -1\\x = \frac {-6 \pm \sqrt {36i ^ 2}} {4}\\x = \frac {-6 \pm i \sqrt {36}} {4}\\x = \frac {-6 \pm6i} {4}$

We have two complex roots:

$x_ {1} = \frac {-6 + 6i} {4} = \frac {-3 + 3i} {2}\\x_ {2} = \frac {-6-6i} {4} = \frac {-3-3i} {2}$

Answer:

$x_ {1} = \frac {-6 + 6i} {4} = \frac {-3 + 3i} {2}\\x_ {2} = \frac {-6-6i} {4} = \frac {-3-3i} {2}$

Answer 2

Answer:

x= (-3+3i ) /2  or x=( -3-3i) / 4

Step-by-step explanation:

Given equation is:

2x²+6x+9=0

ax²+bx+c=0 is general quadratic equation.

x= (-b±√b²-4ac) / 2a is quadratic formula to find the value of x.

comparing given equation with quadratic formula,we get

a= 2  ,b= 6 and c= 9

putting above value in quadratic formula,we get

x= (-6±√6²-4(2)(9)) / 2(2)

x=  (-6±√36-72) / 4

x = (-6±√-36) / 4

x= (-6±√-1√36)  / 4

x= (-6±6i) /4

x= 2(-3±3i)/ 4

x= (-3+3i) /2  or x= (-3-3i) / 4 is solution of 2x²+6x+9=0.