Respuesta :

discipulus

Answer:

[tex] \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} \\ = \frac{ - 2b + 0}{2a} = \frac{ - 2b}{2a} \\ \therefore \: \alpha +\beta = - \frac{b}{a} \\ \alpha \times \beta = \frac{ - b + \sqrt{ {b}^{2} - 4ac} }{2a} \times \frac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a} \\ = \frac{ {b}^{2} - ( {b}^{2} - 4ac) }{4 {a}^{2} } \\ = \frac{4ac}{4 {a}^{2} } \\ \therefore \alpha \times \beta = \frac{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=-\frac{b}{a}\\s=\frac{c}{a}[/tex]

devishri1977

Answer:

Step-by-step 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 = [tex]\frac{-b}{a}[/tex]

2) Product of roots = [tex]\frac{c}{a}[/tex]

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.

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