Polynomials
Grade Class 10
Question:
<p>If sum of all zeros of the polynomial 5x<sup>2</sup> - (3 + k)x + 7 is zero, then zeroes of the polynomial 2x<sup>2</sup> - 2(k+ 11)x + 30 are</p>
<p style="display:inline">7, 9</p>
<p style="display:inline">2, 5</p>
<p style="display:inline">3, 5</p>
<p style="display:inline">3, 6</p>
Step-by-Step Solution
Key Concept: Determine the unknown parameter by applying the sum of zeros formula (-b/a), then substitute this value into the second polynomial and solve for its roots through factorization.
<p>Sum of zeroes of polynomial<br />
5x<sup>2</sup>- (3 + k)x + 7 is<span class="math-tex">$\frac{-[-(3+k)]}{5}$</span>i.e.,<span class="math-tex">$\frac{3+k}{5}$</span><br />
According to question,<span class="math-tex">$\frac{3+k}{5}$</span>= 0<span class="math-tex">$\Rightarrow$</span>k = -3<br />
Now, 2x<sup>2</sup> - 2(k+ 11)x + 30 becomes 2x<sup>2</sup> - 16x + 30.<br />
i.e., 2x<sup>2</sup> - 16x + 30 = 0 or x<sup>2</sup> - 8x + 15 = 0<br />
<span class="math-tex">$\Rightarrow$</span>x = 3, 5<br />
Hence, zeroes of polynomial 2x<sup>2</sup> - 16x + 30 are 3, 5.</p>
Correct Answer: C