using. quadratic equation x^2 +6x +3=0

Question

Please solve

 

 

 

Answer 1

For this, you need to know that the quadratic equation is x=[-b ±√(b²-4ac) ]/ 2a.

 

In the quadratic equation, a b and c refer to the coefficients in front of the variable as well as the number term without the variable. In the equation x²+6x+3=0, a is the number in front of x² which in this case is 1 (there is only 1 x² term). The b is 6 and c is 3.

 

Now it's just a matter of substituting a b and c within the quadratic equation: x= [-6 ± √(6²-4•1•3)] / 2•1 which is the same as x= [-6 ±√(36 -12)] /2. Continuing on, x=[-6 ± √(24)]/2 so one solution is x=(-6 +√24)/2 while the other is x= (-6 –√24)/2.

 

From here, you can split the -6 from √24 from both solutions by making them separate fractions that have the same denominator: x= (-6/2)+(√24/2) and x= (-6/2)-(√24/2). Now the solutions are x= -3 +√24/2 and x= -3 –√24/2 . You can simplify further by taking out the factors that make up 24 (build a factor tree!), so √24 becomes 2√6. 2√6 is being divided by 2 so the 2's cancel leaving √6.

 

Putting it all together, you have the solutions x= -3 + √6 and x= -3 - √6 .

Answer 2

Using quadratic equation 

a = 1

b = 6

c = 3

 

x = (-b ±√(b² - 4ac))/2a = (-6 ±√(36-12))/2 = (-6 ± √24)/2 = -3 ± √6