Question

find the zeros of the quadratic polynomial x^2+x-12 and verify the relationship between zeros and coefficient​

Answer 1

$\large\bf\underline\red{Answer \: :-}$

Given equation : x² + x - 12 = 0

We have to find : zeros of the quadratic equation.

Concept :- We know that the general formula of Quadratic equation is ax² + bx + c = 0, x is variable also zeros and a,b,c are the coefficient. According to the given equation, here, values of a,b,c, is 1 , 1 , -12 are respectively. Here, values of a is not more than 1 because of that we directly factorise the quadratic equation.

Factorisation

$\sf\dashrightarrow{x {}^{2} + x - 12 = 0} \\ \\ \sf\dashrightarrow{x {}^{2} + 4x - 3x - 12 = 0 } \\ \\ \sf\dashrightarrow{x(x + 4) - 3(x + 4) = 0} \\ \\ \sf\dashrightarrow{(x + 4)(x - 3) = 0} \\ \\ \sf\dashrightarrow{x + 4 = 0} \\ \sf\dashrightarrow{x - 3 = 0} \\ \\ \sf\dashrightarrow \pink{x = - 4} \\ \sf\dashrightarrow \pink{x = 3} \: \: \:$

Verification of the zeros

x = -4

$\sf\implies{( - 4) {}^{2} - 4 - 12 = 0} \\ \\ \sf\implies{16 - 4 - 12 = 0} \: \: \: \: \: \: \\ \\ \sf\implies{12 - 12 = 0} \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf\implies \pink{0 = 0} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

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x = 3

$\sf\implies{(3) {}^{2} + 3 - 12 = 0} \\ \\ \sf\implies{9 + 3 - 12 = 0} \: \: \: \: \: \\ \\ \sf\implies{12 - 12 = 0} \: \: \: \: \: \: \: \: \: \: \\ \\ \sf\implies \pink{0 = 0} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

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