Question

if 2/3 and -1/2 are the solutions of the quadratic equation 6x²+ax - b = 0 . Find the value of a and b​

Answer 1

$Sum \: of \: zeroes \: =(α+β)= \frac{2}{ 3 } + \frac{ - 1}{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{4 - 3}{6} = \frac{1}{6} = \frac{ - b}{a} \: \: \: \: \: \: \: \: \: \\ Product\: of \: zeroes = \alpha \beta = \frac{2}{3} \times \frac{ - 1}{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ - 1}{3} = \frac{c}{a} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

For quadratic polynomial

x²-(sum of zeroes)x +(Product of zeroes)=0

x²-(1/6)x+(-1/3)=0

${x}^{2} - \frac{ x}{6} + \frac{ - 1 }{3} = 0 \\ 6 {x}^{2} - x - 2 = 0$

Comparing the given polynomial( 6x²+ax - b = 0) with 6x²-x - 2 = 0, we get

a=-1 and b=2

Answer 2

Answer:

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Step-by-step explanation:

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