Quadratic Equations
Formation of Quadratic Equations
Grade 11
Question:
<p>If complex numbers satisfying <i>α</i> + <i>β</i> = –<i>p</i> and <i>α</i><sup>3</sup> + <i>β</i><sup>3</sup> = <i>q</i>, then a quadratic equation having <i>α</i><sup>3</sup> and <i>β</i><sup>3</sup> as its roots is</p>
<p>(A) <i>(p</i><sup>3</sup> + <i>q)x</i><sup>2</sup> – <i>(p</i><sup>3</sup> + 2<i>q)x</i> + <i>(p</i><sup>3</sup> + <i>q)</i> = 0</p>
<p>(B) <i>(p</i><sup>3</sup> + <i>q)x</i><sup>2</sup> – <i>(p</i><sup>3</sup> – 2<i>q)x</i> + <i>(p</i><sup>3</sup> + <i>q)</i> = 0</p>
<p>(C) <i>(p</i><sup>3</sup> – <i>q)x</i><sup>2</sup> – <i>(5p</i><sup>3</sup> – 2<i>q)x</i> + <i>(p</i><sup>3</sup> – <i>q)</i> = 0</p>
<p>(D) <i>(p</i><sup>3</sup> – <i>q)x</i><sup>2</sup> – <i>(5p</i><sup>3</sup> + 2<i>q)x</i> + <i>(p</i><sup>3</sup> – <i>q)</i> = 0</p>
Step-by-Step Solution
Key Concept: Use sum and product of roots with the identity for sum of cubes to find the quadratic equation whose roots are the cubes of the original roots.
<p>Given: <i>α</i> + <i>β</i> = –<i>p</i> and <i>α</i><sup>3</sup> + <i>β</i><sup>3</sup> = <i>q</i></p><p>For roots <i>α</i><sup>3</sup> and <i>β</i><sup>3</sup>:</p><p>Sum = <i>α</i><sup>3</sup> + <i>β</i><sup>3</sup> = <i>q</i></p><p>Product = <i>α</i><sup>3</sup><i>β</i><sup>3</sup> = (<i>αβ</i>)<sup>3</sup></p><p>From <i>α</i><sup>3</sup> + <i>β</i><sup>3</sup> = (<i>α</i> + <i>β</i>)<sup>3</sup> – 3<i>αβ</i>(<i>α</i> + <i>β</i>), we find <i>αβ</i>.</p><p>∴ Answer is B.</p>
Correct Answer: B