Polynomials
Grade Class 10
Question:
<p>If‘ <span class="math-tex">\(\alpha \)</span> and <span class="math-tex">\(\beta\)</span> are the zeroes of the polynomial 3x<sup>2</sup> + 11x - 4, then the value of <span class="math-tex">\({\alpha ^2} + {\beta ^2}\)</span> is</p>
<p style="display:inline"><span class="math-tex">\(\frac{{145}}{9}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{152}{9}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{{150}}{9}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{{144}}{9}\)</span></p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to determine the sum and product of roots and then substitute them into the algebraic identity alpha^2 + beta^2 = (alpha + beta)^2 - 2*alpha*beta.
<p>Here a = 3, b = 11, c = -4<br />
Since <span class="math-tex">\({\alpha ^2} + {\beta ^2} = {\left( {\alpha + \beta } \right)^2} - 2\alpha \beta \)</span><br />
= <span class="math-tex">\({\left( {\frac{{ - b}}{a}} \right)^2} - 2 \times \frac{c}{a}\)</span> = <span class="math-tex">\(\frac{{{b^2}}}{{{a^2}}} - \frac{{2c}}{a}\)</span> = <span class="math-tex">\(\frac{{{b^2} - 2ac}}{{{a^2}}}\)</span><br />
Putting the values of <span class="math-tex">\(a, \ b\)</span> and <span class="math-tex">\(c,\)</span> we get = <span class="math-tex">\(\frac{{{{\left( {11} \right)}^2} - 2 \times 3 \times \left( { - 4} \right)}}{{{{\left( 3 \right)}^2}}}\)</span><br />
= <span class="math-tex">\(\frac{{121 + 24}}{9}\)</span><br />
= <span class="math-tex">\(\frac{{145}}{9}\)</span></p>
Correct Answer: A