Let and be the roots of the equation x 2 + 2 a x + ( 3 a + 10 ) = 0 such that. Then the set of all possible values of a is : [2026]

Question

Let $α and β$ be the roots of the equation $x^{2} + 2 a x + (3 a + 10) = 0$ such that $α < 1 < β$ . Then the set of all possible values of a is : [2026]

Smart Achievers · Accepted Answer

(1)

$∵$ $α < 1 < β$

$f (1) < 0$

$\Rightarrow 1 + 2 a + (3 a + 10) < 0$

$\Rightarrow 5 a + 11 < 0$

$a < \frac{- 11}{5}$

$∴$ $a \in (- \infty, \frac{- 11}{5})$

Solution

Q.
Let $α and β$ be the roots of the equation $x^{2} + 2 a x + (3 a + 10) = 0$ such that $α < 1 < β$ . Then the set of all possible values of a is : [2026]

1 $(- \infty, \frac{- 11}{5})$

2 $(- \infty, - 3)$

3 $(- \infty, - 2) \cup (5, \infty)$

4 $(- \infty, \frac{- 11}{5}) \cup (5, \infty)$

Ans.
(1)

$∵$ $α < 1 < β$

$f (1) < 0$

$\Rightarrow 1 + 2 a + (3 a + 10) < 0$

$\Rightarrow 5 a + 11 < 0$

$a < \frac{- 11}{5}$

$∴$ $a \in (- \infty, \frac{- 11}{5})$

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Solution

Q. Let α and β be the roots of the equation x2+2ax+(3a+10)=0 such that α<1<β. Then the set of all possible values of a is : [2026]

1 (-∞,-115)

2 (-∞,-3)

3 (-∞,-2)∪(5,∞)

4 (-∞,-115)∪(5,∞)

Ans. (1) ∵ α<1<β f(1)<0 ⇒1+2a+(3a+10)<0 ⇒5a+11<0 a<-115 ∴ a∈(-∞,-115)