Question

quadratic polynomial 6x²+x-12 has zeros as a and ß. now form a quadratic polynomial whose zeroes are 3 a and 3ß​

Answer 1

Answer:

2x^2+x-6=0

Step-by-step explanation:

$6 {x}^{2} + x - 12 \\ ( \div ) \: by \: 6 \\ {x }^{2} + (x \div 6) - 2$

Here,

$\alpha + \beta = - 1 \div 6 \\ \alpha \beta = - 2$

For the new zeros, ( i.e.)

$for \: 3 \alpha \: and \: 3 \beta$

For sum of zeros,

$3 \alpha + 3 \beta \\ = 3( \alpha + \beta ) \\ = 3( - 1 \div 6) \\ = - 1 \div 2$

For product of zeros,

$(3 \alpha )(3 \beta ) \\ = 3( \alpha \beta ) \\ = 3( - 2) \\ = - 6$

Then,

${x}^{2} - (3 \alpha + 3 \beta )x + (3 \alpha )(3 \beta ) = 0$

Therefore,

${x}^{2} - ( - 1 \div 2)x + ( - 6) = 0 \\ ( \times ) \: by \: 2 \\ 2 {x }^{2} + x - 6 = 0$