Quadratic Equations
Roots of Quadratic Equations
Grade 11
Question:
<p>Two non-integer roots of \(\left(\frac{3x-1}{2x+3}\right)^4 - 5\left(\frac{3x-1}{2x+3}\right)^2 + 4 = 0\) are</p>
<p>(a) \(-5/7, -2/5\)</p>
<p>(b) \(-2/5, 7/5\)</p>
<p>(c) \(5/7, 7/5\)</p>
<p>(d) \(-2/5, 3/5\)</p>
Step-by-Step Solution
Key Concept: Use substitution to convert the equation into a quadratic form, then solve for both the substituted variable and the original variable.
<p><strong>Solution:</strong></p><p>Let $\left(\frac{3x-1}{2x+3}\right)^2 = t$</p><p>Then the given equation becomes $t^2 - 5t + 4 = 0$</p><p>$(t-1)(t-4) = 0 \Rightarrow t = 1$ or $t = 4$</p><p>When $t = 1$: $\frac{3x-1}{2x+3} = \pm 1$</p><p>For $\frac{3x-1}{2x+3} = 1$: $3x - 1 = 2x + 3 \Rightarrow x = 4$</p><p>For $\frac{3x-1}{2x+3} = -1$: $3x - 1 = -2x - 3 \Rightarrow 5x = -2 \Rightarrow x = -2/5$</p><p>When $t = 4$: $\frac{3x-1}{2x+3} = \pm 2$</p><p>For $\frac{3x-1}{2x+3} = 2$: $3x - 1 = 4x + 6 \Rightarrow x = -7$</p><p>For $\frac{3x-1}{2x+3} = -2$: $3x - 1 = -4x - 6 \Rightarrow 7x = -5 \Rightarrow x = -5/7$</p><p>The two non-integer roots are $-2/5$ and $-5/7$.</p><p>∴ Answer is (a).</p>
Correct Answer: A