Question

find zeroes of quadratic polynomial 6x²-3-7x and verify the relation between zeroes anf its coefficient

Answer 1

Heya !!
Here is your answer.
Given : 6x² - 3 - 7x

Reordering according to degree of polynomial••

6x² - 7x - 3
Using Hit and Trial Method ::::::
We have to find two numbers a and b such that a+b = -7 and a×b = -18.
Looking at the prime factorisation of 18 = 2×3×3
We get a and b as 2 and 9.

So,
6x² - 7x - 3 = 0
=> 6x² - 9x + 2x - 3 = 0
=> 3x (2x - 3) + 1 (2x - 3) = 0
=> (3x + 1)(2x - 3) = 0

3x + 1 = 0 and 2x - 3 = 0
x = -1/3 and x = 3/2.

Verification :

Comparing the equation with ax² + bx + c = 0,
we get
a = 6, b = -7 and c = -3.

Now,
Sum of zeroes = $\frac{-b}{a}$
Sum of zeroes :
$= > \frac{ - 1}{3} + \frac{3}{2} \\ = > \frac{ - 2 + 9}{6} = \frac{7}{6}$
And,
$\frac{ - b}{a} = - (\frac{ - 7}{6} ) = \frac{7}{6}$
LHS = RHS
Now,
Product of Zeroes = $\frac{c}{a}$
Product of zeroes :
$= > \frac{ - 1}{3} \times \frac{3}{2} = \frac{ - 1}{2}$
And,
$\frac{c}{a} = \frac{3}{6} = \frac{1}{2}$
LHS = RHS

Hence, proved.

Hope You Got It

Answer 2

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

$6 {x}^{2} - 3 - 7x \\ \\ = 6 {x}^{2} + 2x - 9x - 3 \\ \\ = 2x(3x + 1) - (3 x + 1) \\ \\ = (2x - 3)(3x + 1) \\ \\ 2x - 3 = 0 \\ \\ 2x = 3 \\ \\ x = \frac{3}{2} \\ \\ 3x + 1 = 0\\ \\ 3x = - 1 \\ \\ x = \frac{ - 1}{3} \\ \\let \: the \: zeroes \:of \: the \: polynomial \: be \: \alpha \: and \: \beta \\ \\ sum \: of \: zeroes \: ( \alpha \: and \: \beta ) = \frac{ - coefficient \: of \: x}{coefficient \: of \: {x}^{2} } \\ \\ \frac{3}{2} + ( - \frac{1}{3} ) = \frac{ - 7}{6} \\ \\ \frac{9 - 2}{6} = - ( - \frac{7}{6} ) \\ \\ \frac{7}{6} = \frac{7}{6} \\ \\ <strong>Produ</strong><strong>ct</strong>\: <strong>of</strong> \: <strong>zeroes</strong> \: (\alpha \beta ) = \frac{constant \: term \: }{coefficient \: of \: {x}^{2} } \\ \\ \frac{ - 1}{3} \times \frac{3}{2} = \frac{ - 3}{6} \\ \\ \frac{ - 3}{6} = \frac{ - 3}{6}$

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$\Large{\underline{\underline{\bf{Hence\: Verified\: !}}}}$

$\Large{\underline{\underline{\bf{Thanks}}}}$