Answer
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Hint: Any value is said to be a root of the equation only when it results to zero, when substituted in the equation.
Complete step-by-step answer:
It has been given, ${y^2} + 5y + 6 = 0$
Put y = 4 in the equation, we get
16 + 20 + 6 = 42 Since, the remainder for the equation is not zero, hence, 4 is not the root.
Similarly, Put y = -2 in the equation, we get
4 - 10 + 6 = 0 Since, the remainder for the equation is not zero, hence, -2 is the root.
Similarly, Put y = -3 in the equation, we get
9 - 15 + 6 = 0 Since, the remainder for the equation is zero, hence, -3 is the root.
Therefore, the answer is False.
Note: One must also note that it is a quadratic equation, hence it can not have more than 2 roots, so by default the answer was false. One can find the roots of the quadratic equation by using factorisation method or quadratic formula.
Complete step-by-step answer:
It has been given, ${y^2} + 5y + 6 = 0$
Put y = 4 in the equation, we get
16 + 20 + 6 = 42 Since, the remainder for the equation is not zero, hence, 4 is not the root.
Similarly, Put y = -2 in the equation, we get
4 - 10 + 6 = 0 Since, the remainder for the equation is not zero, hence, -2 is the root.
Similarly, Put y = -3 in the equation, we get
9 - 15 + 6 = 0 Since, the remainder for the equation is zero, hence, -3 is the root.
Therefore, the answer is False.
Note: One must also note that it is a quadratic equation, hence it can not have more than 2 roots, so by default the answer was false. One can find the roots of the quadratic equation by using factorisation method or quadratic formula.
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