Question

Find the zeroes of the quadratic polynomial 6x^2+x-2 and verify the relationship between the zeroes and the coefficient ......please give me answer....fastly ​

Answer 1

SOLUTION:-

⠀⠀⠀⠀⠀${\boxed{\blue{\tt{Question }}}}$

• Find the zeroes of the quadratic polynomial 6x^2+x-2 and verify the relationship between the zeroes and the coefficient

⠀⠀⠀⠀⠀${\boxed{\blue{\tt{Given}}}}$

• p(x)= 6x²+x-2

⠀⠀⠀⠀⠀${\boxed{\blue{\tt{To\: Calculate}}}}$

• Zeros of Quadratic polynomial

• And verify relationship

⠀⠀⠀⠀⠀${\boxed{\blue{\tt{Explanation }}}}$

• p(x) = 6x²+x-2=0

✏ 6x²+(4-3)x-2=0

✏ 6x²+4x-3x-2=0

✏ 2x(3x+2)-1(3x+2)=0

✏ (3x+2)(2x-1)=0

✏3x+2=0,2x-1=0

✏3x=-2,2x=1

✏ x= -2/3,x=1/2

Zeros are -2/3 and 1/2

⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀☆ Sum of zeros = -b/a

$➨ \: \frac{ - 2}{3} + \frac{1}{2} = \frac{ - 1}{6}$

Take LCM

$➨ \: \frac{ - 4 + 3}{6} = \frac{ - 1}{6}$

$➨ \: \frac{ - 1}{6} = \frac{ - 1}{6}$

⠀⠀⠀⠀⠀☆ Product of zeros = c/a

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

$➨ \: \frac{ - 2}{6} = \frac{ - 2}{ 6}$

⠀⠀⠀⠀⠀${\huge{\underline{\underline{\sf{\blue{Hence\: Verified !!}}}}}}$

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Answer 2

S O L U T I O N :

We have quadratic polynomial p(x) = 6x² + x - 2 & zero of the polynomial p(x) = 0 .

$\underline{\underline{\tt{Using\:\:by\:\:factorization\:\:method\::}}}$

$\mapsto\sf{6x^{2} + x -2=0}$

$\mapsto\sf{6x^{2} +4x -3x -2=0}$

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

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

$\mapsto\sf{3x+2=0\:\:\:Or\:\:\:2x-1=0}$

$\mapsto\sf{3x=-2\:\:\:Or\:\:\:2x=1}$

$\mapsto\bf{x=-2/3\:\:\:Or\:\:\:x=1/2}$

∴ α = -2/3 & β = 1/2 are the zeroes of the given polynomial .

As we know that given polynomial compared with ax² + bx + c;

• a = 6
• b = 1
• c = -2

Now;

$\underline{\mathcal{SUM\:OF\:THE\:ZEROES\::}}$

$\mapsto\tt{\alpha + \beta =\dfrac{-b}{a} =\bigg\lgroup \dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2}} \bigg\rgroup }$

$\mapsto\tt{\dfrac{-2}{3} + \dfrac{1}{2} =\dfrac{-1}{6}}$

$\mapsto\tt{\dfrac{-4+3}{6} =\dfrac{-1}{6}}$

$\mapsto\bf{\dfrac{-1}{6} =\dfrac{-1}{6}}$

$\underline{\mathcal{PRODUCT\:OF\:THE\:ZEROES\::}}$

$\mapsto\tt{\alpha \times \beta =\dfrac{c}{a} =\bigg\lgroup \dfrac{Constant\:term}{Coefficient\:of\:x^{2}} \bigg\rgroup }$

$\mapsto\tt{\dfrac{-2}{3} \times \dfrac{1}{2} =\dfrac{-2}{6}}$

$\mapsto\tt{\cancel{\dfrac{-2}{6}} =\cancel{\dfrac{-2}{6}}}$

$\mapsto\bf{-1/3 =-1/3}$

Thus;

The relationship between zeroes & coefficient are verified .