Problem 97 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

For what values of \(a\) does the function \(x^{2}+2(a+1) x+9 a-5\) has i. no real roots. ii. only negative roots. iii. only positive roots.

Step-by-Step Solution

Answer

The ranges of \(a\) for which (i) there are no real roots, (ii) there are only negative roots, and (iii) there are only positive roots, depend not only on the discriminant but also the nature of the coefficients such that they can be solved as in steps 2 to 5.

Step 1: Identify the coefficients of x.

Looking at the equation \(x^{2}+2(a+1)x+9a-5=0\), it can be rewritten in the form of \(ax^{2}+bx+c=0\) where \(a=1\), \(b=2(a+1)\), and \(c=9a-5\).

Step 2: Conditions for no real roots.

The condition for having no real roots is that the discriminant is less than zero, i.e. \(b^{2}-4ac < 0\). Substituting the values of \(a\), \(b\), and \(c\), this forms the inequality \([2(a+1)]^{2}-4*1*(9a-5) < 0\). This can be simplified as \(4a^{2}-8a+4 -36a + 20 < 0\). This then simplifies as \(4a^{2} - 44a + 24 < 0\). Dividing by a common value we get \(a^{2}-11a+6 < 0\). Let us solve this later in step 5.

Step 3: Conditions for only negative roots.

For the roots to be negative, the formula \(-b/2a\) must return a negative value as it gives the axis of symmetry (which will correspond to the maximum or minimum point of a parabola given by a quadratic equation). Substituting values, \(-b/2a\) becomes \(-2(a+1)/2*1\), resulting \(-(a+1)\). For \(-(a+1)\) to be negative, \(a\) needs to be less than \(-1\). This is the range of \(a\) to have negative roots.

Step 4: Conditions for only positive roots.

Similarly to step 3, for only positive roots the value of \(-b/2a\) must return a positive value. Substituting values, \(-(a+1)\) should be positive. Therefore, \(a\) should be greater than \(-1\). This is the range of \(a\) to have positive roots.

Step 5: Solve the inequality.

Returning to the inequality from step 2, solve \(a^{2}-11a+6 < 0\). First, solve this as an equality to find critical points, \(a^{2}-11a+6 = 0\). Factorizing, this becomes \((a-2)(a-3) = 0\). The solutions are \(a = 2\) and \(a = 3\). Using these critical points, divide the number line into 3 intervals: \(a < 2\), \(2 < a < 3\), and \(a > 3\). Substituting test points from these intervals into the left hand side of the inequality, find whether each interval satisfies the inequality. The final solution will be the range of \(a\) that satisfies the inequality thus giving no real roots.

Key Concepts

Understanding the Discriminant of Quadratic Equations

The discriminant of a quadratic equation is a powerful tool in algebra that tells us about the nature of the roots the equation possesses. It is found in the standard quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) as the part under the square root, \( b^2 - 4ac \).

When working with the discriminant, there are three possible scenarios:

• If the discriminant is positive (\( b^2 - 4ac > 0 \)), the equation has two distinct real roots.
• If the discriminant is zero (\( b^2 - 4ac = 0 \)), the equation has exactly one real root, also known as a repeated or double root.
• If the discriminant is negative (\( b^2 - 4ac < 0 \)), the equation has no real roots but two complex roots.

In the given exercise, analyzing the discriminant as \( [2(a+1)]^2 - 4*1*(9a-5) < 0 \) gave us insight into the values of \( a \) for which no real roots exist.

Determining Conditions for Negative Quadratic Roots

Negative roots of a quadratic equation are found when the graph of the equation (a parabola) doesn't intersect with the positive side of the x-axis. To know whether a quadratic equation has only negative roots, we look at the vertex of the parabola, since the vertex's x-value (\( -\frac{b}{2a} \) from the vertex formula) will indicate the roots’ sign.

For roots to be strictly negative:

• The vertex must be on the negative x-axis.
• The parabola must open upwards if \( a > 0 \) and downwards if \( a < 0 \).

In our case, since \( a = 1 \) is positive, and the vertex is \( -(a+1) \) which is on the negative x-axis if \( a < -1 \) – it guarantees that both roots of the equation are negative.

Conditions for Positive Quadratic Roots

On the flip side, positive roots occur when the parabola intersects the x-axis only in the positive region. As explained for negative roots, the vertex of the parabola provides key information.

For a quadratic equation to have positive roots, the vertex (\( -\frac{b}{2a} \) again) should be positive, which will be the case when \( a+1 > 0 \) or \( a > -1 \). Provided the parabola opens upwards, which is true when the leading coefficient \( a \) is positive, the roots will be solely in the positive x-axis region. This establishes \( a > -1 \) as the criterion for the given function to have positive roots.