Find the sum and product of roots of the quadratic equations given below1.x²-5x+6=02.x²-6=03.3x²+7x=2x-54.3x²-7x+6=65.x²+5x+1=3x²+6​

Step-by-step explanation:Sum: $$5$$5, Product: $$6$$6Sum: $$0$$0, Product: $$6$$6Sum: $$-\frac{1}{3}$$− 31 , Product: $$1$$1Sum: $$-\frac{5}{3}$$− 35 , Product: $$0$$0Sum: $$\frac{7}{3}$$37 , Product: $$2$$2For the equation $$x^{2}-5x+6=0$$x 2 −5x+6=0, the sum of the roots is the negative of the coefficient of $$x$$x, which is $$5$$5, and the product of the roots is the constant term, which is $$6$$6For the equation $$x^{2}-6=0$$x 2 −6=0, the sum of the roots is $$0$$0 (since the coefficient of $$x$$x is $$0$$0) and the product of the roots is the constant term, which is $$6$$6For the equation $$3y^{2}+y+1=0$$3y 2 +y+1=0, the sum of the roots is $$-\frac{1}{3}$$− 31 (since the coefficient of $$y$$y is $$1$$1) and the product of the roots is $$1$$1 (since the constant term is $$1$$1).For the equation $$3x^{2}+7x=2x-5$$3x 2 +7x=2x−5, first rearrange the equation to standard form: $$3x^{2}+5x+5=0$$3x 2 +5x+5=0. The sum of the roots is $$-\frac{5}{3}$$− 35 (since the coefficient of $$x$$x is $$5$$5) and the product of the roots is $$5$$5 (since the constant term is $$5$$5).For the equation $$3x^{2}-7x+6=6$$3x 2 −7x+6=6, first subtract $$6$$6 from both sides to get $$3x^{2}-7x=0$$3x 2 −7x=0. This can be factored as $$x(3x-7)=0$$x(3x−7)=0. The sum of the roots is $$0$$0 (since one root is $$0$$0) and the product of the roots is $$0$$0 (since one root is $$0$$0).So, the answers are:Sum: $$5$$5, Product: $$6$$6Sum: $$0$$0, Product: $$6$$6Sum: $$-\frac{1}{3}$$− 31 , Product: $$1$$1Sum: $$-\frac{5}{3}$$− 35 , Product: $$0$$0Sum: $$\frac{7}{3}$$37 , Product: $$2$$2

Answered by cyrusvillanueva14yro | 2025-07-30