Magic Math Solver | 10th Grade Problems

Problem 1: Solving Equations

(i) \(\sin^2 x - 5 \sin x + 4 = 0\)
(ii) \(12x^3 + 8x = 29x^2 - 4\)

(i) Solution for \(\sin^2 x - 5 \sin x + 4 = 0\):

1 Let \( y = \sin x \). The equation becomes \( y^2 - 5y + 4 = 0 \)

2 Factor the quadratic: \( (y - 1)(y - 4) = 0 \)

3 Solutions: \( y = 1 \) or \( y = 4 \)

4 But \( \sin x \) cannot be 4 (since range of sine is [-1,1]), so only \( \sin x = 1 \)

5 Therefore, \( x = \frac{\pi}{2} + 2\pi n \), where \( n \) is any integer

Final Answer: \( x = \frac{\pi}{2} + 2\pi n \), \( n \in \mathbb{Z} \)

(ii) Solution for \(12x^3 + 8x = 29x^2 - 4\):

1 Rearrange: \(12x^3 - 29x^2 + 8x + 4 = 0\)

2 Try rational roots: \( x = 2 \) works as \( 12(8) - 29(4) + 16 + 4 = 96 - 116 + 20 = 0 \)

3 Perform polynomial division or factor: \( (x-2)(12x^2 - 5x - 2) = 0 \)

4 Factor the quadratic: \( (3x-2)(4x+1) = 0 \)

5 Solutions: \( x = 2 \), \( x = \frac{2}{3} \), \( x = -\frac{1}{4} \)

Final Answer: \( x = -\frac{1}{4} \), \( x = \frac{2}{3} \), or \( x = 2 \)

Problem 2: Rational Roots

(i) \(2x^3 - x^2 - 1 = 0\)
(ii) \(x^8 - 3x + 1 = 0\)

(i) Solution for \(2x^3 - x^2 - 1 = 0\):

1 Possible rational roots: \( \pm1, \pm\frac{1}{2} \) (factors of constant over factors of leading coefficient)

2 Test \( x = 1 \): \( 2 - 1 - 1 = 0 \) → It's a root!

3 Factor: \( (x-1)(2x^2 + x + 1) = 0 \)

4 Quadratic has discriminant \( D = 1 - 8 = -7 \) → no real roots

Final Answer: Only rational root is \( x = 1 \)

(ii) Solution for \(x^8 - 3x + 1 = 0\):

1 Possible rational roots: \( \pm1 \) only

2 Test \( x = 1 \): \( 1 - 3 + 1 = -1 \neq 0 \)

3 Test \( x = -1 \): \( 1 + 3 + 1 = 5 \neq 0 \)

Final Answer: No rational roots exist for this equation

Problem 3: Exponential Equation

Solve: \(8x^{\frac{3}{2n}} - 8x^{\frac{-3}{2n}} = 63\)

1 Let \( y = x^{\frac{3}{2n}} \). Then \( x^{\frac{-3}{2n}} = \frac{1}{y} \)

2 Equation becomes: \( 8y - \frac{8}{y} = 63 \)

3 Multiply through by y: \( 8y^2 - 63y - 8 = 0 \)

4 Solve quadratic: \( y = \frac{63 \pm \sqrt{3969 + 256}}{16} = \frac{63 \pm 65}{16} \)

5 Solutions: \( y = 8 \) or \( y = -\frac{1}{8} \)

6 Since \( y = x^{\frac{3}{2n}} \) must be positive, take \( y = 8 \)

7 Then \( x^{\frac{3}{2n}} = 8 \) → \( x = 8^{\frac{2n}{3}} = (2^3)^{\frac{2n}{3}} = 2^{2n} \)

Final Answer: \( x = 4^n \) (since \( 2^{2n} = (2^2)^n = 4^n \))

Problem 4: Radical Equation

Solve: \(2\sqrt{\frac{x}{a}} + 3\sqrt{\frac{a}{x}} = \frac{b}{a} + \frac{6a}{b}\)

1 Let \( y = \sqrt{\frac{x}{a}} \), then \( \sqrt{\frac{a}{x}} = \frac{1}{y} \)

2 Equation becomes: \( 2y + \frac{3}{y} = \frac{b}{a} + \frac{6a}{b} \)

3 Notice the right side can be written as \( \frac{b}{a} + 6\left(\frac{a}{b}\right) \)

4 Let \( k = \frac{b}{a} \), then equation is \( 2y + \frac{3}{y} = k + \frac{6}{k} \)

5 Compare terms: \( 2y = k \) and \( \frac{3}{y} = \frac{6}{k} \)

6 From \( 2y = k \) and \( \frac{3}{y} = \frac{6}{k} \), substitute \( k = 2y \) into second equation

7 \( \frac{3}{y} = \frac{6}{2y} \) → \( \frac{3}{y} = \frac{3}{y} \) (always true)

8 So \( y \) can be any positive real number, and \( k = 2y \)

9 Recall \( y = \sqrt{\frac{x}{a}} \), so \( x = a y^2 \)

Final Answer: \( x = a y^2 \) where \( y \) is any positive real number, and \( \frac{b}{a} = 2y \)

Problem 5: Polynomial Equations

(i) \(6x^4 - 35x^3 + 62x^2 - 35x + 6 = 0\)
(ii) \(x^4 + 3x^3 - 3x - 1 = 0\)

(i) Solution for \(6x^4 - 35x^3 + 62x^2 - 35x + 6 = 0\):

1 This is a reciprocal equation. Divide by \( x^2 \): \( 6x^2 - 35x + 62 - \frac{35}{x} + \frac{6}{x^2} = 0 \)

2 Group terms: \( 6(x^2 + \frac{1}{x^2}) - 35(x + \frac{1}{x}) + 62 = 0 \)

3 Let \( y = x + \frac{1}{x} \), then \( x^2 + \frac{1}{x^2} = y^2 - 2 \)

4 Substitute: \( 6(y^2 - 2) - 35y + 62 = 0 \) → \( 6y^2 - 35y + 50 = 0 \)

5 Solve quadratic: \( y = \frac{35 \pm \sqrt{1225-1200}}{12} = \frac{35 \pm 5}{12} \)

6 Solutions: \( y = \frac{10}{3} \) or \( y = \frac{5}{2} \)

7 For \( y = \frac{10}{3} \): \( x + \frac{1}{x} = \frac{10}{3} \) → \( 3x^2 - 10x + 3 = 0 \)

8 Solutions: \( x = 3 \) or \( x = \frac{1}{3} \)

9 For \( y = \frac{5}{2} \): \( x + \frac{1}{x} = \frac{5}{2} \) → \( 2x^2 - 5x + 2 = 0 \)

10 Solutions: \( x = 2 \) or \( x = \frac{1}{2} \)

Final Answer: \( x = \frac{1}{3} \), \( x = \frac{1}{2} \), \( x = 2 \), or \( x = 3 \)

(ii) Solution for \(x^4 + 3x^3 - 3x - 1 = 0\):

1 Factor by grouping: \( x^4 - 1 + 3x^3 - 3x = 0 \)

2 \( (x^2 - 1)(x^2 + 1) + 3x(x^2 - 1) = 0 \)

3 Common factor: \( (x^2 - 1)(x^2 + 1 + 3x) = 0 \)

4 First factor: \( x^2 - 1 = 0 \) → \( x = \pm 1 \)

5 Second factor: \( x^2 + 3x + 1 = 0 \) → \( x = \frac{-3 \pm \sqrt{9 - 4}}{2} = \frac{-3 \pm \sqrt{5}}{2} \)

Final Answer: \( x = -1 \), \( x = 1 \), or \( x = \frac{-3 \pm \sqrt{5}}{2} \)

Problem 6: Exponential Equation

Find all real numbers satisfying \(4^x - 3(2^{x+2}) + 2^5 = 0\)

1 Rewrite terms: \(4^x = (2^2)^x = 2^{2x}\), and \(2^{x+2} = 4 \cdot 2^x\), \(2^5 = 32\)

2 Equation becomes: \(2^{2x} - 12 \cdot 2^x + 32 = 0\)

3 Let \( y = 2^x \), then equation is \( y^2 - 12y + 32 = 0 \)

4 Solve quadratic: \( y = \frac{12 \pm \sqrt{144 - 128}}{2} = \frac{12 \pm 4}{2} \)

5 Solutions: \( y = 8 \) or \( y = 4 \)

6 For \( y = 8 \): \( 2^x = 8 \) → \( x = 3 \)

7 For \( y = 4 \): \( 2^x = 4 \) → \( x = 2 \)

Final Answer: \( x = 2 \) or \( x = 3 \)

Problem 7: Polynomial Equation

Solve the equation \(6x^4 - 5x^3 - 38x^2 - 5x + 6 = 0\) if it is known that \(\frac{1}{3}\) is a solution.

1 Since \( x = \frac{1}{3} \) is a root, \( (3x - 1) \) is a factor

2 Perform polynomial division or factor: \( (3x - 1)(2x^3 - x^2 - 13x - 6) = 0 \)

3 Try \( x = -2 \) in the cubic: \( -16 - 4 + 26 - 6 = 0 \) → \( (x + 2) \) is a factor

4 Factor further: \( (3x - 1)(x + 2)(2x^2 - 5x - 3) = 0 \)

5 Factor the quadratic: \( (2x + 1)(x - 3) = 0 \)

6 All factors: \( (3x - 1)(x + 2)(2x + 1)(x - 3) = 0 \)

Final Answer: \( x = -2 \), \( x = -\frac{1}{2} \), \( x = \frac{1}{3} \), or \( x = 3 \)