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Mathematics

If the equation 2x2 - 6x + p = 0 has real and different roots, then the values of p are given by

  1. p<92p \lt \dfrac{9}{2}

  2. p≤92p \le \dfrac{9}{2}

  3. p>92p \gt \dfrac{9}{2}

  4. p≥92p \ge \dfrac{9}{2}

Quadratic Equations

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Answer

Given ,

2x2 - 6x + p = 0 has real and different roots

Comparing equation with ax2 + bx + c = 0
a= 2 , b = -6 , c = p

Since, equation has real and different roots

∴ b2 - 4ac > 0

⇒(−6)2−4×2×p>0⇒(−6)2−8p>0⇒36−8p>0⇒8p<36⇒p<368p<92\Rightarrow (-6)^2 - 4 \times 2 \times p \gt 0 \\[0.5em] \Rightarrow (-6)^2 - 8p \gt 0 \\[0.5em] \Rightarrow 36 - 8p \gt 0 \\[0.5em] \Rightarrow 8p \lt 36 \\[0.5em] \Rightarrow p \lt \dfrac{36}{8} \\[0.5em] p \lt \dfrac{9}{2}

∴ Option 1 is the correct option.

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