Question

Q1In a quadratic equation ax2 + bx+c=0
(a) Sum of the roots =
(b) Product of the roots =

Answer 1

Step-by-step explanation:

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,

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

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,

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

Now,

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

And,

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

Answer 2

1. the sum of its roots :-. -b/a
2. product of the roots :-. c/a quadratic equation may be expressed as a product of two binomials . here ,a and b are called the roots of the given quadratic equation.