The quadratic equation 2x^2 + 6x + 7 = 0 has roots A and B. Write down the value of A + B and the value of AB

Upon being told to find the roots of a quadratic we should investigate the various different ways of finding these roots. With experience you will learn to be able to find roots without many calculations usually if the roots are integers. However, this isn't the case for this question so we have two primary methods of finding these roots. We can use the quadratic formula or 'complete the square'. One method will suffice but I'll show you both and the second can be used as a method to check your answer, something I'd recommend doing if you have time left over in the exam. Firstly, using the quadratic formula we wish to have the quadratic equation in the form ax^2 + bx + c = 0. Conveniently, this equation is already in that form so we can now move on to equate the coefficients of the x^2 value, the coefficients of the x value and the remaining constant. This gives us: a = 2, b = 6 and c = 7. We can now implement it into the formula (-b +or- sqrt(b^2 - 4ac))/2a. Since we have a +or- there is a possibility of two different roots if the determinant (b^2 - 4ac) equals zero (a likely scenario considering the question uses two different letters for the possible roots). Substituting a, b and c into the equation we have (-6 +or- sqrt(6^2 - 4(2)(7)))/(2(2)). Simplifying this we have -3/2 +or- sqrt(-20)/4. Since there is no real value for the square of a negative number we separate sqrt(-20) into (sqrt(-1))(sqrt(20)) using the laws for surd numbers. We can now replace sqrt(-1) with the imaginary number i and proceed with simplifying the expression making sure not to combine the coefficients of the real and imaginary values. Our two remaining roots the real value plus the positive and negative values of the imaginary number. In other words, A = -3/2 + sqrt(5)/2 and B = -3/2 - sqrt(5)/2. All that is left is to find their sum and their product. In finding their sum we add the real values and add the imaginary values making sure not to mix the two. Our imaginary values cancel and we are left with A + B = -3/2 + sqrt(5)/2 i + -3/2 - sqrt(5)/2 i = -3, our first answer. The product works similarly, we have AB = (-3/2 + sqrt(5)/2)*(-3/2 - sqrt(5)/2). Although it is not essential to notice this we see we have a difference of two squares here so we have 9/4 - (5/4i^2). Since i=sqrt(-1), i^2 = -1. We can now write our equation as 9/4 + 5/4 = 3.5, our second answer. The second way of finding the roots of a quadratic is by completing the square as follows: Our first step is to check does the coefficient of x^2 equal one. Since it does not we must divide our entire equation by this constant (2) such that we have x^2 on its own. x^2 + 3x + 7/2 = 0. We want to rearrange the equation to be in the form (x + a)^2 + k = 0 where a and k are constants. We can do so by adding the square of half of the x coefficient to both sides of the equals (make sure to include the minus if required). In other words: x^2 + 3x + 7/2 = 0 => x^2 + 3x + 7/2 + 9/4 = 9/4. We can now factorise the left hand side and get (x + 3/2)^2 + 7/2 = 9/4. Moving the constants to the right hand side of the equals we get (x+3/2)^2 = -5/4. Getting the square root of both sides we have x + 3/2 = +or- sqrt(-5/4). (Be very careful not to forget the +or- required because both positive and negative values produce a positive value when squared as this is a very common error). Subtract 3/2 from both sides and convert sqrt(-5/4) to sqrt(5)/2 i using the same rules as above. We see that we have the same two roots as we found with the quadratic formula checking we are correct. Find AB and A + B the same way as above. I want to stress that I am aware I was being quite slow making sure that all the major steps and assumptions were covered. When I see you are capable of doing most of this without all of these steps written down there won't be a need for me to include all these steps.