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If p ,q in {1,2,3,4,5} , then find the n...

If `p ,q in {1,2,3,4,5}` , then find the number of equations of form `p^2x^2+q^2x+1=0` having real roots.

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To solve the problem of finding the number of equations of the form \( p^2x^2 + q^2x + 1 = 0 \) that have real roots, we need to analyze the conditions under which the quadratic equation has real roots. The condition for a quadratic equation \( ax^2 + bx + c = 0 \) to have real roots is that its discriminant \( D \) must be greater than or equal to zero. 1. **Identify the coefficients**: - Here, \( a = p^2 \), \( b = q^2 \), and \( c = 1 \). 2. **Calculate the discriminant**: \[ D = b^2 - 4ac = (q^2)^2 - 4(p^2)(1) = q^4 - 4p^2 \] 3. **Set the discriminant condition**: For the quadratic to have real roots, we need: \[ D \geq 0 \implies q^4 - 4p^2 \geq 0 \implies q^4 \geq 4p^2 \] 4. **Rearranging the inequality**: This can be rearranged to: \[ q^2 \geq 2p \] 5. **Evaluate the possible values of \( p \) and \( q \)**: Since \( p \) and \( q \) can take values from the set \( \{1, 2, 3, 4, 5\} \), we will check each possible value of \( q \) and find the corresponding values of \( p \) that satisfy the inequality \( q^2 \geq 2p \). - **For \( q = 1 \)**: \[ q^2 = 1 \implies 1 \geq 2p \implies p \leq 0.5 \quad \text{(no valid } p\text{)} \] Valid \( p \): 0 - **For \( q = 2 \)**: \[ q^2 = 4 \implies 4 \geq 2p \implies p \leq 2 \quad \text{(valid } p: 1, 2\text{)} \] Valid \( p \): 2 - **For \( q = 3 \)**: \[ q^2 = 9 \implies 9 \geq 2p \implies p \leq 4.5 \quad \text{(valid } p: 1, 2, 3, 4\text{)} \] Valid \( p \): 4 - **For \( q = 4 \)**: \[ q^2 = 16 \implies 16 \geq 2p \implies p \leq 8 \quad \text{(valid } p: 1, 2, 3, 4, 5\text{)} \] Valid \( p \): 5 - **For \( q = 5 \)**: \[ q^2 = 25 \implies 25 \geq 2p \implies p \leq 12.5 \quad \text{(valid } p: 1, 2, 3, 4, 5\text{)} \] Valid \( p \): 5 6. **Count the total valid combinations**: - For \( q = 1 \): 0 possibilities - For \( q = 2 \): 2 possibilities - For \( q = 3 \): 4 possibilities - For \( q = 4 \): 5 possibilities - For \( q = 5 \): 5 possibilities Total = \( 0 + 2 + 4 + 5 + 5 = 16 \) Thus, the total number of equations of the form \( p^2x^2 + q^2x + 1 = 0 \) that have real roots is **16**.

To solve the problem of finding the number of equations of the form \( p^2x^2 + q^2x + 1 = 0 \) that have real roots, we need to analyze the conditions under which the quadratic equation has real roots. The condition for a quadratic equation \( ax^2 + bx + c = 0 \) to have real roots is that its discriminant \( D \) must be greater than or equal to zero. 1. **Identify the coefficients**: - Here, \( a = p^2 \), \( b = q^2 \), and \( c = 1 \). 2. **Calculate the discriminant**: \[ D = b^2 - 4ac = (q^2)^2 - 4(p^2)(1) = q^4 - 4p^2 ...
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