Question

Find the zeros of the quadratic polynomial 6x^2 -3 -7x and verify the relationship between

the zeros and the coefficients.​

Answer 1

Question:

Find the zeros of the quadratic polynomial 6x² -7x -3 and verify the relationship between the zeros and the coefficients.

To find:

★ To verify the relationship between the zeros and the coefficients.

Answer:

$\star \underline{ \: \bold{ (\alpha + \beta ) =\sf \pink{ \dfrac{7}{6}} } \: }$

$\star \underline{ \: \bold{ ( \alpha \beta ) =\sf \pink{ \dfrac{ - 1}{2}} } \: }$

Given:

★ An equation is given, ${6x}^{2}-7x-3 = 0$

Step-by-step explanation:

Now factorization,

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

Product :- - 18 = - 9 × 2

Sum :- - 7 = ( - 9 ) + ( 2 )

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

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

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

$\implies 2x - 3 = 0 \: \: , \: \: 3x + 1 = 0$

$\implies 2x = 3 \: \: , \: \: 3x = - 1$

$\implies \underline{x = \dfrac{3}{2} \: \: , \: \: x = \dfrac{ - 1}{3} }$

$\therefore \boxed{ \bold{\alpha = \frac{3}{2} \: and \: \beta = \frac{ - 1}{3} } }$

Now verification :-

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

$Here\: a=6,\:b=-7,\:c=-3$

We know that,

$\boxed{Sum \: of \: roots = \dfrac{ - b}{a} }$

$\hookrightarrow ( \alpha + \beta ) = \dfrac{ - b}{a}$

$\hookrightarrow \dfrac{3}{2} + \big(\dfrac{ - 1}{3} \big) = \dfrac{ - ( - 7)}{6}$

$\hookrightarrow \dfrac{9 - 2}{6} = \dfrac{7}{6}$

$\hookrightarrow \boxed{ \dfrac{7}{6} = \dfrac{7}{6} }$

$\underline{ \: \bold{ (\alpha + \beta ) = \dfrac{7}{6} = \dfrac{7}{6} } \: }$

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$\boxed{Product \: of \: roots = \dfrac{ c}{a} }$

$\rightsquigarrow ( \alpha \beta ) = \dfrac{c}{a}$

$\rightsquigarrow \dfrac{ \not{3}}{2} \times \dfrac{ - 1}{ \not{3}} = \dfrac{ - 3}{6}$

$\rightsquigarrow \boxed{ \dfrac{ - 1}{2} = \dfrac{ - 1}{2} }$

$\underline{ \: \bold{ ( \alpha \beta ) = \dfrac{ - 1}{2} = \dfrac{ - 1}{2} } \: }$

Hence verified .

Formula used:

$\bigstar \: Sum \: of \: roots = \dfrac{ - b}{a}$

$\bigstar \: Product \: of \: roots = \dfrac{c}{a}$