Question

Find the zeros of the following quadratic polynomials by factorization method and verify the relation between the zeros and tne coefficients: 6x^2 + x - 12

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Answer 1

Step-by-step explanation:

${6x}^{2} + x - 12 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - 12 \times 6 = - 72 \\ by \: factorization \: \: \: \: 9 \times - 8 = - 72. \: \: \: 9 + (- 8 )= 1 \\ {6x}^{2} + 9x - 8x - 12 = 0 \\ ( {6x}^{2} + 9x)( - 8x - 12) = 0 \\ 3x(2x + 3) - 4(2 + 3) = 0 \\ (3x - 4)(2x + 3) = 0 \\3x - 4=0 \: or \: 2x + 3= 0 \\ 3x = 4 \: \: \: or \: \: \: 2x = - 3 \\ x = \frac{4}{3} \: \: \: or \: \: \: x = \frac{ - 3}{2}$

We find the factors are 9 & -8 of quadratic polynomial

${6x}^{2} + x - 12$

Let's change the sign of 9 & -8

Therefore -9 & 8

Let divide - 9 & 8 by coefficient of x^2

Then answer is -3/2 & 4/3

It is relation between zeros & coefficient

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