If $6$ lies between the roots of the equation $x ^ { 2 } + 2 ( a - 3 ) x + 9 = 0$ then
- $a \in \left( - \dfrac { 3 } { 4 } , \infty \right)$
- $a \in \left( - \infty , - \dfrac { 3 } { 4 } \right)$
- $a \in ( - \infty , 0 ) \cup ( 6 , \infty )$
- $a \in \left( - \dfrac { 3 } { 4 } , 6 \right)$
Reveal answer
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B
Correct answer
Explanation
For a quadratic f(x) = x^2 + 2(a-3)x + 9, if 6 lies between the roots, then f(6) < 0. f(6) = 36 + 12(a-3) + 9 = 36 + 12a - 36 + 9 = 12a + 9. Setting 12a + 9 < 0 gives 12a < -9, so a < -3/4.
AI explanation
For the value 6 to lie between the roots of the quadratic equation x^2 + 2(a - 3)x + 9 = 0, the condition f(6) < 0 must be met while the parabola opens upwards. Substituting x = 6 into the equation yields 6^2 + 2(a - 3)(6) + 9 < 0. This simplifies to 36 + 12a - 36 + 9 < 0, which reduces to 12a + 9 < 0. Solving this linear inequality for a gives a < -9/12, or a < -3/4. The correct result is a belongs to the interval from negative infinity to negative three-fourths.