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Find the values of pp that will make the quadratic (5p−1)x2−4x+(2p−1)(5p - 1)x^2 - 4x + (2p - 1) a perfect square.

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Full solution

Q. Find the values of pp that will make the quadratic (5p−1)x2−4x+(2p−1)(5p - 1)x^2 - 4x + (2p - 1) a perfect square.
  1. Identify Form: Identify the general form of a perfect square quadratic.\newlineA perfect square quadratic is of the form (ax+b)2=ax2+2abx+b2(ax + b)^2 = ax^2 + 2abx + b^2.
  2. Compare with General Form: Compare the given quadratic with the general form.\newlineThe given quadratic is 5p - 1)x^2 - 4x + (2p - 1)\. We need to find the value of \$p such that the quadratic becomes a perfect square.
  3. Determine Coefficient of xx: Determine the coefficient of xx in the perfect square. For the quadratic to be a perfect square, the linear coefficient (2ab)(2ab) must be equal to −4-4. In our case, a=5p−1a = \sqrt{5p - 1} and bb would be the square root of the constant term, which we will find later.
  4. Solve for b: Solve for b in terms of pp. We have 2ab=−42ab = -4, so 2×5p−1×b=−42 \times \sqrt{5p - 1} \times b = -4. Solving for b gives us b=−42×5p−1b = \frac{-4}{2 \times \sqrt{5p - 1}}.
  5. Express as b2b^2: Express the constant term as b2b^2. The constant term in the perfect square, b2b^2, must be equal to (2p−1)(2p - 1). So, (−4/(2∗5p−1))2=2p−1(-4 / (2 * \sqrt{5p - 1}))^2 = 2p - 1.
  6. Simplify to Solve for p: Simplify the equation to solve for p. (−4/(2∗5p−1))2=2p−1(-4 / (2 * \sqrt{5p - 1}))^2 = 2p - 1 simplifies to (16/(4∗(5p−1)))=2p−1(16 / (4 * (5p - 1))) = 2p - 1. Further simplification gives us 4/(5p−1)=2p−14 / (5p - 1) = 2p - 1.
  7. Clear Denominator: Multiply both sides by (5p−1)(5p - 1) to clear the denominator.4=(2p−1)(5p−1)4 = (2p - 1)(5p - 1). This expands to 4=10p2−5p−2p+14 = 10p^2 - 5p - 2p + 1.
  8. Combine and Solve for pp: Combine like terms and bring all terms to one side to solve for pp.10p2−7p−3=010p^2 - 7p - 3 = 0.
  9. Factor Quadratic Equation: Factor the quadratic equation to find the values of pp. We need to find two numbers that multiply to (10×−3)=−30(10 \times -3) = -30 and add up to −7-7. These numbers are −10-10 and 33. However, the quadratic does not factor easily, so we will use the quadratic formula instead.
  10. Apply Quadratic Formula: Apply the quadratic formula to find pp.p=−(−7)±(−7)2−4⋅10⋅(−3)2⋅10p = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \cdot 10 \cdot (-3)}}{2 \cdot 10}.
  11. Simplify Quadratic Formula: Simplify the quadratic formula.\newlinep=7±49+12020p = \frac{7 \pm \sqrt{49 + 120}}{20}.\newlinep=7±16920p = \frac{7 \pm \sqrt{169}}{20}.\newlinep=7±1320p = \frac{7 \pm 13}{20}.
  12. Find Possible Values for p: Find the two possible values for pp.p=7+1320p = \frac{7 + 13}{20} or p=7−1320p = \frac{7 - 13}{20}.p=2020p = \frac{20}{20} or p=−620p = \frac{-6}{20}.p=1p = 1 or p=−310p = -\frac{3}{10}.

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