Quadratic Equations and Theory of Equations — Flashcards
Telangana Open School (TOSS) · Class 12 · Mathematics
23 flashcards for Quadratic Equations and Theory of Equations (Telangana Open School (TOSS) Class 12 Mathematics) to test yourself on key terms and facts.
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Solve by factorization: \( 6x^2 + 5x - 6 = 0 \)
Answer
Step 1: Split middle term: \( 6x^2 + 9x - 4x - 6 = 0 \) Step 2: Factor by grouping: \( 3x(2x + 3) - 2(2x + 3) = 0 \) Step 3: Factor common binomial: \( (2x + 3)(3x - 2) = 0 \) Step 4: Set each factor …
Solve: \( 3\sqrt{2}x^2 + 7x - 3\sqrt{2} = 0 \) by factorization
Answer
Step 1: Split middle term: \( 3\sqrt{2}x^2 + 9x - 2x - 3\sqrt{2} = 0 \) Step 2: Group terms: \( 3x(\sqrt{2}x + 3) - \sqrt{2}(\sqrt{2}x + 3) = 0 \) Step 3: Factor common binomial: \( (\sqrt{2}x + 3)(3x…
Solve: \( (a + b)^2x^2 + 6(a^2 - b^2)x + 9(a - b)^2 = 0 \)
Answer
Step 1: Recognize as perfect square trinomial. Step 2: Rewrite: \( [(a + b)x + 3(a - b)]^2 = 0 \) Step 3: Solve: \( (a + b)x + 3(a - b) = 0 \) Step 4: \( x = \frac{-3(a - b)}{a + b} = \frac{3(b - a)}{…
Solve using quadratic formula: \( 2x^2 - 3x + 3 = 0 \)
Answer
Step 1: Identify \( a = 2, b = -3, c = 3 \) Step 2: Compute discriminant: \( D = b^2 - 4ac = (-3)^2 - 4(2)(3) = 9 - 24 = -15 \) Step 3: Since \( D < 0 \), roots are complex: \( x = \frac{-(-3) \pm \sq…
Solve: \( -x^2 + \sqrt{2}x - 1 = 0 \)
Answer
Step 1: Multiply by -1: \( x^2 - \sqrt{2}x + 1 = 0 \) Step 2: Use formula: \( a = 1, b = -\sqrt{2}, c = 1 \) Step 3: \( D = (\sqrt{2})^2 - 4(1)(1) = 2 - 4 = -2 \) Step 4: Roots: \( x = \frac{\sqrt{2} …
For what value of \( k \) does \( (4k + 1)x^2 + (k + 1)x + 1 = 0 \) have equal roots?
Answer
Step 1: For equal roots, \( D = 0 \) Step 2: \( a = 4k + 1, b = k + 1, c = 1 \) Step 3: \( D = (k + 1)^2 - 4(4k + 1)(1) = 0 \) Step 4: Expand: \( k^2 + 2k + 1 - 16k - 4 = 0 \Rightarrow k^2 - 14k - 3 =…
If \( \alpha, \beta \) are roots of \( 3x^2 - 5x + 9 = 0 \), find \( \alpha^2 + \beta^2 \)
Answer
Step 1: Use identity: \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \) Step 2: From equation: \( \alpha + \beta = \frac{5}{3}, \alpha\beta = \frac{9}{3} = 3 \) Step 3: \( \alpha^2 + \beta^…
If \( \alpha, \beta \) are roots of \( 3y^2 + 4y + 1 = 0 \), form equation with roots \( \alpha^2, \beta^2 \)
Answer
Step 1: \( \alpha + \beta = -\frac{4}{3}, \alpha\beta = \frac{1}{3} \) Step 2: Sum of new roots: \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(-\frac{4}{3}\right)^2 - 2\left(\frac{…
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