Question

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Answer 1

`\alpha = \frac{ - b + \sqrt{ {b}^(2) - 4ac} }{2a} \\ \beta= \frac{ - b - \sqrt{ {b}^(2) - 4ac} }{2a} \\ \alpha +\beta = \frac{ - b + \sqrt{ {b}^(2) - 4ac} }{2a} + \frac{ - b - \sqrt{ {b}^(2) - 4ac} }{2a} \\ = ( - 2b + 0)/(2a) = ( - 2b)/(2a) \\ \therefore \: \alpha +\beta = - (b)/(a) \\ \alpha * \beta = \frac{ - b + \sqrt{ {b}^(2) - 4ac} }{2a} * \frac{ - b - \sqrt{ {b}^(2) - 4ac} }{2a} \\ = \frac{ {b}^(2) - ( {b}^(2) - 4ac) }{4 {a}^(2) } \\ = \frac{4ac}{4 {a}^(2) } \\ \therefore \alpha * \beta = (c)/(a) \\ (x - \alpha)(x - \beta) = 0 \\ {x }^(2) + (\alpha + \beta)x + \alpha\beta = 0\\\\Let \:D =\sqrt{ {b}^(2) - 4ac}\\ If \:D >0 \Rightarrow \alpha, \beta \in \mathbb R\:(2 \: real \:roots)\\ If D =0 \Rightarrow \alpha= \beta \in \mathbb R\: (one \:real\: root) \\ If D <0 \Rightarrow \alpha, \beta \in \mathbb C\:(complex\: roots)\\At least 4examples\\{(x-1)}^(2)=0\\{x}^(2)-x-1=0 \\{x}^(2)-2x+13=0\\7{x}^(2)-49x-7=0\\ Applications\\ut+g{t}^(2)=s\\a=g\\u=-(b)/(a)\\s=(c)/(a)`

Answer 2

Explanation:

1) Determining the nature of roots:

Quadratic equation: ax² + bx +c = 0

Find D (Discriminant) using the D = b² - 4ac

If D = 0 , then the two roots are equal.

If D > 0, then the equation has two distinct real roots.

If D < 0, then the roots are complex/ imaginary.

One can find the sum and product of the roots using the below mentioned formula.

1) sum of roots =
`(-b)/(a)`

2) Product of roots =
`(c)/(a)`

If sum and product of the roots are given, one can frame the quadratic equation by x² - (sum of roots)*x + product of roots = 0

2) i) x² - 6x + 9 = 0

a = 1 ; b = -6 and c = 9

D = (-6)² - 4 * 1 * 9 = 36 - 36 = 0

D = 0. So, two equal roots

ii) x² - 4x + 3 = 0

a = 1 ; b = -4 ; c = 3

D = (-4)² - 4 * 1 * 3 = 16 - 12 = 4

D > 0. So, two distinct real roots.

iii) x² - 4x - 3 = 0

a = 1 ; b = -4 ; c = -3

D = (-4)² - 4*1*(-3) = 16 + 12 = 28 > 0

D > 0. So, two distinct real roots.

iv) x² - 4x + 7 = 0

a = 1 ; b = -4 ; c = 7

D =(-4)² - 4 *1 *7 = 16 - 28 = - 12

D< 0. So, two imaginary/ complex roots.