Maths Q10 – AP-EAMCET 2026

Home Maths Complex & QuadraticQuestion #60

The number of integral values of 'm' for which the quadratic expression

(1 + 2 m) x^{2} - 2 (1 + 3 m) x + 4 (1 + m), x \in R

is always positive, is

Clue 1 of 3: Quadratic Expression Positivity

For a quadratic expression $A x^{2} + B x + C$ to be always positive for all real values of $x$ , two fundamental conditions must be satisfied: the leading coefficient must be positive, and the discriminant must be negative.

🧠 Full Solution Path

Step 1: Identify Conditions for Always Positive Quadratic Expression✦ Active

For a quadratic expression $P (x) = a x^{2} + b x + c$ to be always positive for all real values of $x$ (i.e., $P (x) > 0$ ), two conditions must be met simultaneously:

1. The coefficient of $x^{2}$ must be positive: $a > 0$ .

2. The discriminant must be negative: $D = b^{2} - 4 a c < 0$ .

For the given quadratic expression, $(1 + 2 m) x^{2} - 2 (1 + 3 m) x + 4 (1 + m)$ , we identify the coefficients:

a = (1 + 2 m)

b = - 2 (1 + 3 m)

c = 4 (1 + m)

💡 Teacher's Secret Hint

Remember that these conditions are crucial for a quadratic to maintain a constant sign over its entire domain. If $a < 0$ and $D < 0$ , the expression would be always negative.

Step 2: Apply the First Condition (

a > 0

)○ Expand

Apply the condition that the coefficient of $x^{2}$ must be positive:

1 + 2 m > 0

2 m > - 1

m > - \frac{1}{2}

m > - 0.5

This gives us the first range for $m$ .

💡 Teacher's Secret Hint

Carefully handle inequalities, especially when multiplying or dividing by negative numbers (though not applicable here).

Step 3: Apply the Second Condition (

D < 0

)○ Expand

Calculate the discriminant $D = b^{2} - 4 a c$ and set it less than zero:

D = [- 2 (1 + 3 m)]^{2} - 4 (1 + 2 m) (4 (1 + m))

D = 4 (1 + 3 m)^{2} - 16 (1 + 2 m) (1 + m)

Now, set $D < 0$ and simplify:

4 (1 + 3 m)^{2} - 16 (1 + 2 m) (1 + m) < 0

Divide by 4:

(1 + 3 m)^{2} - 4 (1 + 2 m) (1 + m) < 0

Expand the terms:

(1 + 6 m + 9 m^{2}) - 4 (1 + m + 2 m + 2 m^{2}) < 0

(1 + 6 m + 9 m^{2}) - 4 (1 + 3 m + 2 m^{2}) < 0

1 + 6 m + 9 m^{2} - 4 - 12 m - 8 m^{2} < 0

Combine like terms:

m^{2} - 6 m - 3 < 0

💡 Teacher's Secret Hint

Be careful with algebraic expansion and sign changes. A common mistake is distributing the negative sign incorrectly.

Step 4: Solve the Quadratic Inequality for

m

○ Expand

To solve $m^{2} - 6 m - 3 < 0$ , first find the roots of the quadratic equation $m^{2} - 6 m - 3 = 0$ using the quadratic formula $m = \frac{- B \pm \sqrt{B^{2} - 4 A C}}{2 A}$ :

m = \frac{- (- 6) \pm \sqrt{(- 6)^{2} - 4 (1) (- 3)}}{2 (1)}

m = \frac{6 \pm \sqrt{36 + 12}}{2}

m = \frac{6 \pm \sqrt{48}}{2}

m = \frac{6 \pm 4 \sqrt{3}}{2}

m = 3 \pm 2 \sqrt{3}

Approximate values: \sqrt{3} \approx 1.732

2 \sqrt{3} \approx 3.464

m_{1} = 3 - 2 \sqrt{3} \approx 3 - 3.464 = - 0.464

m_{2} = 3 + 2 \sqrt{3} \approx 3 + 3.464 = 6.464

Since the quadratic $m^{2} - 6 m - 3$ opens upwards (coefficient of $m^{2}$ is positive), $m^{2} - 6 m - 3 < 0$ means $m$ is between its roots:

3 - 2 \sqrt{3} < m < 3 + 2 \sqrt{3}

Approximately: - 0.464 < m < 6.464

💡 Teacher's Secret Hint

Remember that for a quadratic $A x^{2} + B x + C$ , if $A > 0$ , then $A x^{2} + B x + C < 0$ implies $x$ lies between the roots. If $A < 0$ , $A x^{2} + B x + C < 0$ implies $x$ lies outside the roots.

Step 5: Combine Conditions and Count Integral Values○ Expand

We must satisfy both conditions simultaneously:

1. From Step 2: $m > - 0.5$

2. From Step 4: $- 0.464 < m < 6.464$

The intersection of these two conditions is: $- 0.464 < m < 6.464$ .

We need to find the number of *integral values* of $m$ in this interval. The integers that satisfy this range are:

0, 1, 2, 3, 4, 5, 6

Counting these values, there are 7 integral values of $m$ .

💡 Teacher's Secret Hint

Always check the boundaries of the interval carefully when determining integral values. Integers must be strictly greater than -0.464 and strictly less than 6.464.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Maths Question 10 – AP-EAMCET 2026

Clue 1 of 3: Quadratic Expression Positivity

🧠 Full Solution Path