Question

Find the zeroes of the given polynomial:​

Answer 1

★ The zeroes of the given polynomial are –1/3 and 3/2.

Step-by-step explanation:

Given -

‎ ‎ ‎ ‎We've been given a polynomial and have been asked to find the zeroes of the polynomial and verify the relationship between zeroes and their coefficients. The given polynomial is: 6x² – 3 – 7x.

‎

Solution -

‎ ‎ ‎ ‎Determining the zeroes of the polynomial:-

$\dashrightarrow{ \sf{ {6x}^{2} - 3 - 7x }} \\ \\ \dashrightarrow{ \sf{ {6x}^{2} - 7x - 3 = 0 }} \\ \\ \dashrightarrow{ \sf{ {6x}^{2} + 2x - 9x - 3 = 0 }} \\ \\ \dashrightarrow{ \sf{2x(3x + 1) - 3(3x + 1) = 0}} \\ \\ \dashrightarrow{ \sf{(2x - 3)(3x + 1) = 0}} \\ \\ \dashrightarrow \underline{ \boxed{ \frak{ \red{ \pmb{x = \dfrac{3}{2}}}} \: \sf{ and} \: \frak{ \pmb{ \red{x = \frac{ - 1}{3} }}}}}$

Therefore, the zeroes of the given polynomial are –1/3 and 3/2.

‎

Verification -

‎ ‎ ‎ ‎ ‎In any quadratic equations (ax² + bx + c = 0), if α and β are its roots, then sum and product is given by:

Sum of zeroes:

‎

$\dashrightarrow{ \sf{( \alpha + \beta ) = \dfrac{ - (Coefficient \: of \: x)}{(Coefficient \: of \: {x}^{2} )} }} \\ \\ \\ \dashrightarrow{ \sf{ \bigg( \dfrac{ - 1}{3} \bigg) + \bigg( \dfrac{3}{2} \bigg) = \dfrac{ - ( - 7)}{6} }} \\ \\ \\ \dashrightarrow{ \sf{ \dfrac{7}{6} = \dfrac{ - ( - 7)}{6} }} \\ \\ \\ \dashrightarrow \underline{ \boxed{ \frak{ \pmb{ \dfrac{7}{6} = \dfrac{7}{6} }}}}$

‎

‎

Product of zeroes:

‎

$\dashrightarrow{ \sf{( \alpha \times \beta ) = \dfrac{(Constant \: term)}{(Coefficient \: of \: {x}^{2}) } }} \\ \\ \\ \dashrightarrow{ \sf{ \bigg( \dfrac{ - 1}{3} \bigg) \times \bigg( \dfrac{3}{2} \bigg) = \dfrac{ - 3}{6} }} \\ \\ \\ \dashrightarrow{ \sf{ \dfrac{ - 3}{6} = \dfrac{ - 3}{6} }} \\ \\ \\ \dashrightarrow \underline{ \boxed{ \frak{ \pmb{ \dfrac{ - 1}{2} = \dfrac{ - 1}{2} }}}}$

‎

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎Hence, verified!

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Answer 2

★ Concept :

Here we have a quadratic polynomial (6x² - 3 - 7x) , we had to find the zeroes of the given polynomial and verify the relationship between zeroes and their coefficients.

Let's do it !!

ㅤㅤㅤㅤㅤㅤㅤ━━━━━━━━

$\underbrace{\underline{\sf{Required\:solution\::}}}$

✎ Using middle splitting method to find ㅤthe zeroes of the polynomial :

$\qquad:\implies\:\sf 6x^2 - 7x - 3 = 0$

$\qquad:\implies\:\sf 6x^2 - 9x + 2x - 3 = 0$

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

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

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

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

$\qquad:\implies\:{\bold {x = {\red{\dfrac{-1}{3}}\:,\:x = {\red{\dfrac{3}{2}}}}}}$

$\small\underline{\boxed{\tt{The\:zeroes(\alpha\:,\:\beta)\:of\:the\:given\:polynomial\:=\:\rm\purple{\dfrac{-1}{3}}\:,\:\rm\purple{\dfrac{3}{2}}}}}$

ㅤㅤㅤㅤㅤㅤㅤ━━━━━━━━

✎ Verifying relationship between ㅤㅤㅤㅤzeroes and their coefficients :

$\qquad:\implies\:\sf Sum\:of\:zeroes\:=\:\red{\dfrac{-b}{a}}$

★ Values that we have :

» b = Coefficient of x = -7

» a = Coefficient of x² = 6

✎ Putting all values :

$\qquad:\implies\:\sf \alpha\:+\:\beta\:=\:\dfrac{-(-7)}{6}$

$\qquad:\implies\:\sf \dfrac{-1}{3}\:+\:\dfrac{3}{2} =\: \dfrac{7}{6}$

$\qquad:\implies\:\sf \dfrac{-2 + 9}{6} =\: \dfrac{7}{6}$

$\qquad:\implies\:\sf \dfrac{7}{6} =\: \dfrac{7}{6}$

ㅤㅤㅤㅤㅤL.H.S = R.H.S

$\qquad:\implies\:\sf Product\:of\:zeroes\:=\:\red{\dfrac{c}{a}}$

★ Values that we have :

» c = Constant term = -3

» a = Coefficient of x² = 6

✎ Putting all values :

$\qquad:\implies\:\sf \alpha\:\times\:\beta\:=\:\dfrac{-3}{6}$

$\qquad:\implies\:\sf \dfrac{-1}{3}\:\times\:\dfrac{3}{2}\:=\:\dfrac{-3}{6}$

$\qquad:\implies\:\sf \dfrac{-3}{6}\:=\:\dfrac{-3}{6}$

$\qquad:\implies\:\sf {\dfrac{\cancel{-3}}{\cancel{6}}}\:=\:{\dfrac{\cancel{-3}}{\cancel{6}}}$

$\qquad:\implies\:\sf \dfrac{-1}{2}\:=\:\dfrac{-1}{2}$

ㅤㅤㅤㅤㅤL.H.S = R.H.S

$\qquad\qquad\underline{\tt{\purple\ddag\:Hence\:Verified\:\purple\ddag}}$

Solution completed !!