lf $a x^{2} - (2 a + 3) x + (3 + 5 a) = 0$ has no real roots, then $a$ lies in the inverval

(- \frac{3}{4}, \frac{3}{4})

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(- \infty, \frac{3}{4})

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(- \infty, - \frac{3}{4})

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(- \infty, - \frac{3}{4}) \cup (\frac{3}{4}, \infty)

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Solution

The correct option is D $(- \infty, - \frac{3}{4}) \cup (\frac{3}{4}, \infty)$
If $a x^{2} + b x + c$ has no real roots, then
$\Rightarrow b^{2} - 4 a c < 0$
$\Rightarrow (2 a + 3)^{2} - 4 (a) (3 + 5 a) < 0.$
$\Rightarrow 4 a^{2} + 9 + 12 a - 12 a - 20 a^{2} < 0$
$\Rightarrow 9 - 16 a^{2} < 0$
$\Rightarrow (- \infty, - \frac{3}{4}) \cup (\frac{3}{4}, \infty)$

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lf ax2−(2a+3)x+(3+5a)=0 has no real roots, then a lies in the inverval

The correct option is D (−∞,−34)∪(34, ∞)If ax2+bx+c has no real roots, then⇒b2−4ac<0⇒(2a+3)2−4(a)(3+5a)<0.⇒4a2+9+12a−12a−20a2<0⇒9−16a2<0⇒(−∞,−34)∪(34, ∞)

lf $a x^{2} - (2 a + 3) x + (3 + 5 a) = 0$ has no real roots, then $a$ lies in the inverval