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### High School Mathematics6.10 Solving Equations Reducible to Quadratic Form

 Example: Solve, (x2 + 2x)2 - 11(x2 + 2x) + 24 = 0 Solution: Put x2 + 2x = a in the given equation, we get a2 - 11a + 24 = 0 a2 - 11a + 24 = a2 - 8a - 3a + 24 = a(a - 8) - 3(a - 8) = (a - 8)(a - 3) a = 3 or 8 Case (i) a = 3, x2 + 2x = 3 x2 + 2x - 3 = 0 x2 + 3x - x - 3 = 0 x(x + 3) - (x + 3) = 0 (x + 3)(x - 1) x = -3 and 1 Case (ii) a = 8, x2 + 2x = 8 x2 + 2x - 8 = 0 x2 + 4x - 2x - 8 = 0 x(x + 4) - 2(x + 4) = 0 (x + 4)(x - 2) = 0 x = -4 and 2 Therefore, the roots of the given equation are -3, 1, -4 and 2 Directions: Solve the following problems. Also write at least 5 examples of your own.
 Q 1: Solve, 6x4 + 11x2 - 3 = 0.Ö(-3/2) and Ö(-1/3)Ö(3/2) and Ö(-1/3)Ö(-3/2) and Ö(--1/3) Q 2: Solve, (x2 - 2x)2 - 11(x2 - 2x) + 24 = 0.-1, 2, 3 and 4-1, 2, -3 and 4-1, -2, 3 and 4 Q 3: Solve, 9x4 - 3x2 - 2 = 0.Ö(1/3) and Ö(2/3)Ö(-1/3) and Ö(-2/3)Ö(-1/3) and Ö(2/3) Q 4: Solve, 12x4 + 10x2 + 2 = 0Ö(1/2) and Ö(-1/3)Ö(-1/2) and Ö(--1/3)Ö(-1/2) and Ö(-1/3) Q 5: Solve, 15x4 + 13x2 + 2 = 0.Ö(-2/3) and Ö(-1/5)Ö(-2/3) and Ö(1/5)Ö(2/3) and Ö(-1/5) Q 6: Solve, (x2 - x)2 - 8(x2 - x) + 12 = 0.-1, -2, 2 and 3-1, 2, -2 and -31, -2, 2 and 3 Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only!

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