How do I solve this and then express the sum and the product of the roots?

The question says: Solve the following equation. Then express the sum and the product of the roots to check your solutions.





x^2+6x-7=0








a) x=1; x= -7


sum = 6; prod= -7





b) x= -1; x= 7


sum = -6; prod= -7





c) x=1; x= -7


sum = -6; prod= -7





d) x=1; x= 7


sum =8; prod=7

How do I solve this and then express the sum and the product of the roots?
You need not do any calculations seperately for finding the sum and product


of the roots.


The coefficient of x with opposite sign is the sum of the roots and "c" with the same sign is the product of the roots in case of a quadratic equation ax^2+bx+c=0,when a=1


x^2+6x-7=0


=%26gt;(x-1)(x+7)=0


the roots of x are 1 and -7.OK?


Now look at the original equation.The coefficient of x is +6.Therefore,the sum of the roots must be -6.Check it up


Again look at the original equation.The "C" in our equation is -7.So -7 must be the product of the roots.Check it up and be happy.
Reply:Given a second order equation, if it has roots, of a general form ax^2+bx+c=0, always x1+x2=b/a and x1*x2= -c/a.
Reply:x^2 + 6x - 7 = 0





x^2 + 7x - x - 7 = 0





x (x + 7) - 1 (x + 7) = 0





( x + 7) (x - 1) = 0





x = - 7 or x = 1





Sum -%26gt; -6





Product -%26gt; -7





ie Option C
Reply:The easiest way to solve this would be the quadratic formula:





If ax^2 + bx + c = 0





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


(sqrt = square root)





This will give you two roots (one each for the +/- square root) if (b^2 - 4ac) is positive, which in this case it is. Then just add the two roots together to get the sum, and multiply them together to get the product.





Alternatively, you could factor the equation to get (x + p)(x + q) = 0, where p and q are numbers. Then your roots would be x = -p and x = -q (in other words, the values of x that could make the equation equal 0).