Polynomials
Grade Class 10

Question:

<p>If &alpha;, &beta; are the zeros of the polynomial&nbsp;<em>p</em>(<em>x</em>) = 4<em>x</em><sup>2</sup>&nbsp;+ 3<em>x</em>&nbsp;+ 7, then&nbsp;<span class="math-tex">\(\frac{1}{\alpha}+\frac{1}{\beta}\)</span>&nbsp;is equal to</p>
<p style="display:inline"><span class="math-tex">\(-\frac{7}{3}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{7}{3}\)</span></p>
<p style="display:inline"><span class="math-tex">\(-\frac{3}{7}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{3}{7}\)</span></p>

Step-by-Step Solution

Key Concept: Evaluate symmetric expressions of roots by rewriting them in terms of the sum (-b/a) and product (c/a) of the zeros.
<p>Since&nbsp;<span class="math-tex">\(\alpha\)</span>&nbsp;and&nbsp;<span class="math-tex">\(\beta\)</span>&nbsp;are the zeros of the quadratic polynomial&nbsp;<span class="math-tex">\(p(x)=4 x^{2}+3 x+7\)</span></p> <p><span class="math-tex">\(\alpha+\beta=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}\)</span>&nbsp;<span class="math-tex">\(=\frac{-3}{4}\)</span></p> <p><span class="math-tex">\(\alpha \beta=\frac{\text { Constant term }}{\text { coefficient of } x^{2}}\)</span>&nbsp;&nbsp;<span class="math-tex">\(=\frac{7}{4}\)</span></p> <p>Now,&nbsp;<span class="math-tex">\(\frac{1}{\alpha}+\frac{1}{\beta}\)</span>&nbsp;<span class="math-tex">\(=\frac{\beta+\alpha}{\alpha \beta}\)</span>&nbsp;<span class="math-tex">\(=\frac{\frac{-3}{4}}{\frac{7}{4}}\)</span>&nbsp;<span class="math-tex">\(=\frac{-3}{4} \times \frac{4}{7}\)</span>&nbsp;<span class="math-tex">\(=\frac{-3}{7}\)</span><br /> Thus, the&nbsp;value of&nbsp;<span class="math-tex">\(\frac{1}{a}+\frac{1}{\beta}\)</span>&nbsp;is&nbsp;<span class="math-tex">\(\frac{-3}{7}\)</span>.</p>
Correct Answer: C

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