Find the interval in which a lies such that the roots of the equation (a+1)x2−3ax+4a, is greater than 1.

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granfury90210birm · Accepted Answer

To find the values of a such that real roots of the equation
$(a + 1) x^{2} - 3 a x + 4 a = 0$
are greater than 1.
First the discriminant must be positive
${(3 a)}^{2} - 16 a (a + 1) \geq 0 \to - \frac{16}{7} \leq a \leq 0 (1)$
Second, if we want both roots greater than 1, then the sum must be greater than 2
The sum of the roots is $x_{1} + x_{2} = \frac{3 a}{a + 1}$
$x_{1} + x_{2} > 2 \to \frac{3 a}{a + 1} > 2 \to a < - 1 l or a > 2 (2)$
Putting together (1) and (2) we have that the roots of the given equation are greater than 1 if $- \frac{16}{7} \leq a < - 1$ .

Jameson Jensen · Accepted Answer

Note that solving 3a2(a+1)&amp;lt;1 gets you to (-1,2). In this region the - solution cannot be bigger than 1 since you are subtracting a positive number, so you have to exclude that region

Find the interval in which a lies such

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