Question

In a quadratic equation ax2 + bx+c=0
(a) Sum of the roots =
(b) Product of the roots =

Answer 1

Answer:

$\large\boxed{x=-\frac{c}{2a+b}; a\neq-\frac{b}{2}   }$

Step-by-step explanation:

To solve this problem, first you have to use quadratic equation from left to right. Remember to solve this problem, you had to solve with distributive property and isolate by the c from one sides of the equation.

Given:

ax²+bx+c=0

Solutions:

Sum of the roots is -b/a

Product of the roots is c/a.

First, you have to use quadratic equation formula.

$\large\boxed{\textnormal{Quadratic Equation Formula}}$

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}$

Isolate by the c on one side of the equation.

Subtract c from both sides.

$\displaystyle ax*2+bx+c-c=0-c$

Solve.

$\displaystyle ax*2+bx=-c$

Factor the term of ax*2+bx and try using the distributive property.

$\large\boxed{\textnormal{Distributive Property}}$

$\displaystyle a(b+c)=ab+ac$

$\displaystyle ax*2+bx$

Common term of x.

$\displaystyle x(2a+b)$

Rewrite the whole problem down.

$\displaystyle x(2a+b)=-c$

Then, you divide by 2a+b from both sides.

$\displaystyle \frac{x(2a+b)}{2a+b}=\frac{-c}{2a+b}; a\neq-\frac{b}{2}$

Solve.

$\displaystyle \frac{x(2a+b)}{2a+b}=\frac{-c}{2a+b}; a\neq-\frac{b}{2}=\boxed{x=-\frac{c}{2a+b};a\neq-\frac{b}{2}   }$

As a result, the correct answer is x=-c/2a+b; and a≠-b/2.

Answer 2

Answer: -b/a and c/a

Step-by-step explanation:

Given,

ax² + bx + c  = 0

Dividing by a

x² + (b/a)x + c/a = 0/a

x² + (b/a)x + c/a = 0

x² + (b/a)x  = -c/a

Adding  on (b/2a)² both sides

x² + b²/4a + (b/a)x = -c/a + b²/4a²

x² + b²/4a + (b/a)x = (b²-4ac)/4a²

x² + (b/2a)² + 2×(x)×(b/2a) =  (b²-4ac)/4a²

(x + b/2a)² = (b²-4ac)/4a²

Taking square root of  both sides,

$x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}\\\;\\x=\pm\frac{\sqrt{b^2-4ac}}{2a}-\frac{b}{2a}\\\;\\x=\pm\frac{\sqrt{b^2-4ac}-b}{2a}\\\;\\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

From here we can obtain two roots of the given equation by takin positive and negative sign respectively. If the first root is  α and second is β. We have,

$\alpha=\frac{-b+\sqrt{b^2-4ac}}{2a}\\\;\\\text{and,}\;\;\beta=\frac{-b-\sqrt{b^2-4ac}}{2a}$

Now,

$\text{Sum of roots}=\alpha+\beta\\\;\\=\frac{-b+\sqrt{b^2-4ac}}{2a}+\frac{-b-\sqrt{b^2-4ac}}{2a}\\\;\\=\frac{-b-b}{2a}\\\;\\=\frac{-b}{a}$

And,

$\text{Product of roots}=\alpha\times\beta\\\;\\=\frac{-b+\sqrt{b^2-4ac}}{2a}\;\times\;\frac{-b-\sqrt{b^2-4ac}}{2a}\\\;\\=\frac{(-b)^2-(\sqrt{b^2-4ac})^2}{4a^2}\\\;\\=\frac{b^2-b^2+4ac}{4a^2}\\\;\\=\frac{4ac}{4a^2}\\\;\\=\frac{c}{a}$