Question

Find zeros of the following quadratic polynomial and verify the relationship between the zeros and its coefficients.

$\mathtt{f(x) = {6x}^{2} - 3 - 7x }$
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Answer 1

Answer:

$\huge\mathtt\blue{SOLUTION:-}$

$\mathtt{f(x) = {6x}^{2} - 3 - 7x }$

$\mathtt{ {6x}^{2} - 3 - 7x = 0 }$

$\mathtt{ {6x}^{2} - 7x - 3 = 0 }$

$\mathtt{ {6x}^{2} - 9x + 2x - 3 = 0}$

$\mathtt{ {6x}^{2} + 2x - 9x - 3 = 0 }$

$\mathtt{2x(3x + 1) - 3(3x + 1) = 0}$

$\mathtt{(3x + 1) - (2x - 3) = 0}$

$\mathtt{3x + 1 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 2x - 3 = 0}$

$\mathtt{3x = - 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 2x = 3}$

$\mathtt{x = \frac{ - 1}{3} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x = \frac{3}{2} }$

$\mathtt{ \alpha = \frac{3}{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \beta = \frac{ - 1}{3} }$

$\mathtt \blue{a = 6 \: \: \: \: \: \: \: \: \: \: \: b = - 7 \: \: \: \: \: \: \: \: \: \: \: c = - 3}$

$\mathtt{SUM \: OF \: ZEROS}$

$\mathtt \blue{ \alpha + \beta = \frac{ - b}{a} }$

$\mathtt \blue{ \alpha + \beta = \frac{ - 7}{6} }$

$\mathtt \blue{ \frac{3}{2} + \frac{( - 1)}{3} = \frac{ - ( - 7)}{6} }$

$\mathtt \blue{ \frac{9 - 2}{6} = \frac{7}{6} }$

$\mathtt \blue{ \frac{7}{6} = \frac{7}{6} }$

$\mathtt{LHS=RHS}$

$\mathtt{PRODUCT \: OF \: ZEROS}$

$\mathtt \blue{ \alpha \beta = \frac{c}{a} }$

$\mathtt \blue{ \frac{3}{2} \times \frac{( - 1)}{3} = \frac{ - 3}{6} }$

$\mathtt \blue{ \frac{ - 3}{6} = \frac{ - 3}{6} }$

$\mathtt{LHS=RHS}$

Answer 2

Answer:

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