Question

find the zeros of the quadratic polynomial p(x)-6x2-3-7x​

Answer 1

Step-by-step explanation:

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Answer 2

$\huge\sf\underline{\underline{\pink{ Answer:-}}}$

$zeros = > \frac{3}{2} \: ,\frac{ - 1}{3}$

$\Large{\underline{\underline{\bf{QuEsTiOn:-}}}}$

find the zeros of the quadratic polynomial

$6x {}^{2} - 3 - 7x$

$\sf\underline{\underline{\green{TO \: FIND \: }}}$

Zeros of the quadratic polynomial ?

$\huge\underline\mathbb{\red S\pink {0} \purple {L} \blue {UT} \orange {1}\green {ON :}}$

$= > 6x {}^{2} - 3 - 7x$

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

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

$= > 2x(3x + 1) \: - 3(3x + 1) = 0$

$= >( 2x - 3)(3x + 1) = 0$

$= > Zeros =  \frac{3}{2} , \frac{3}{ - 1} </p><p>$

$verify -$

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

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

$= > \frac{3}{2} - \frac{1}{3} = \frac{ 7}{6}$

$= > \frac{7}{6} = \frac{7}{6}$

RHS = LHS

$= > \alpha \beta = \frac{c}{a}$

$= > \frac{3}{2} \times \frac{ - 1}{3} = \frac{ - 3}{6}$

$= > \frac{ - 3} {6} = \frac{ - 3}{6}$

$= > \frac{ - 1}{2} = \frac{ - 1}{2}$

RHS= LHS

$\sf\underline{\underline{\green{HENCE, \: }}}$

$zeros = \frac{3}{2}, \frac{ - 1}{3}$