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Online Quiz (Worksheet A B C D)

Questions Per Quiz = 2 4 6 8 10

High School Mathematics
6.10 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 - x2 - 2 = 0.
(-1/2) and (2/3)
(-1/2) and (-2/3)
(1/2) and (2/3)

Q 2: Solve, 15x4 + 13x2 + 2 = 0.
(2/3) and (-1/5)
(-2/3) and (-1/5)
(-2/3) and (1/5)

Q 3: 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 4: Solve, (x2 - x)2 - 8(x2 - x) + 12 = 0.
1, -2, 2 and 3
-1, 2, -2 and -3
-1, -2, 2 and 3

Q 5: Solve, (2x2 - 3x)2 - 4(2x2 - 3x) - 5 = 0.
-1, 1, -1/2 and 5/2
-1, 1, 1/2 and 5/2
1, 1, 1/2 and 5/2

Q 6: Solve, 6x4 + 11x2 - 3 = 0.
(-3/2) and (--1/3)
(3/2) and (-1/3)
(-3/2) and (-1/3)

Question 7: This question is available to subscribers only!

Question 8: This question is available to subscribers only!


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