Polynomials | Exercise 2.2

Question 2

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i) 14,−1\dfrac{1}{4}, -1

(ii) 2,13\sqrt{2}, \dfrac{1}{3}

(iii) 0,50, \sqrt{5}

(iv) 1,11, 1

(v) −14,14-\dfrac{1}{4}, \dfrac{1}{4}

(vi) 4,14, 1

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Solution
Understand the Question
  • A quadratic polynomial with given sum of zeroes (SS) and product of zeroes (PP) is given by: P(x)=k[x2−(sum of zeroes)x+(product of zeroes)]=k(x2−Sx+P)P(x) = k\left[x^2 - (\text{sum of zeroes})x + (\text{product of zeroes})\right] = k\left(x^2 - Sx + P\right) where kk is any non-zero real constant.
  • We substitute the given sum and product into the formula and choose a suitable value of kk (typically the denominator) to eliminate fractions.

(i) 14,−1\dfrac{1}{4}, -1

Step 1 · Form the Quadratic Polynomial

Given Sum of zeroes (S)=14\text{Sum of zeroes } (S) = \dfrac{1}{4} and Product of zeroes (P)=−1\text{Product of zeroes } (P) = -1.

The general form of the quadratic polynomial is

P(x)=x2−(sum)x+(product)=x2−(14)x+(−1)=x2−14x−1\begin{aligned} P(x) &= x^2 - (\text{sum})x + (\text{product}) \\[0.6em] &= x^2 - \left(\dfrac{1}{4}\right)x + (-1) \\[0.6em] &= x^2 - \dfrac{1}{4}x - 1 \end{aligned}

Multiplying by 44 to clear the denominator

P(x)=4(x2−14x−1)=4x2−x−4\begin{aligned} P(x) &= 4\left(x^2 - \dfrac{1}{4}x - 1\right) \\[0.6em] &= 4x^2 - x - 4 \end{aligned}
Answer

(i) 4x2−x−44x^2 - x - 4

(ii) 2,13\sqrt{2}, \dfrac{1}{3}

Step 1 · Form the Quadratic Polynomial

Given Sum of zeroes (S)=2\text{Sum of zeroes } (S) = \sqrt{2} and Product of zeroes (P)=13\text{Product of zeroes } (P) = \dfrac{1}{3}.

The general form of the quadratic polynomial is

P(x)=x2−(sum)x+(product)=x2−(2)x+(13)=x2−2x+13\begin{aligned} P(x) &= x^2 - (\text{sum})x + (\text{product}) \\[0.6em] &= x^2 - (\sqrt{2})x + \left(\dfrac{1}{3}\right) \\[0.6em] &= x^2 - \sqrt{2}x + \dfrac{1}{3} \end{aligned}

Multiplying by 33 to clear the denominator

P(x)=3(x2−2x+13)=3x2−32x+1\begin{aligned} P(x) &= 3\left(x^2 - \sqrt{2}x + \dfrac{1}{3}\right) \\[0.6em] &= 3x^2 - 3\sqrt{2}x + 1 \end{aligned}
Answer

(ii) 3x2−32x+13x^2 - 3\sqrt{2}x + 1

(iii) 0,50, \sqrt{5}

Step 1 · Form the Quadratic Polynomial

Given Sum of zeroes (S)=0\text{Sum of zeroes } (S) = 0 and Product of zeroes (P)=5\text{Product of zeroes } (P) = \sqrt{5}.

The general form of the quadratic polynomial is

P(x)=x2−(sum)x+(product)=x2−(0)x+(5)=x2+5\begin{aligned} P(x) &= x^2 - (\text{sum})x + (\text{product}) \\[0.6em] &= x^2 - (0)x + (\sqrt{5}) \\[0.6em] &= x^2 + \sqrt{5} \end{aligned}
Answer

(iii) x2+5x^2 + \sqrt{5}

(iv) 1,11, 1

Step 1 · Form the Quadratic Polynomial

Given Sum of zeroes (S)=1\text{Sum of zeroes } (S) = 1 and Product of zeroes (P)=1\text{Product of zeroes } (P) = 1.

The general form of the quadratic polynomial is

P(x)=x2−(sum)x+(product)=x2−(1)x+(1)=x2−x+1\begin{aligned} P(x) &= x^2 - (\text{sum})x + (\text{product}) \\[0.6em] &= x^2 - (1)x + (1) \\[0.6em] &= x^2 - x + 1 \end{aligned}
Answer

(iv) x2−x+1x^2 - x + 1

(v) −14,14-\dfrac{1}{4}, \dfrac{1}{4}

Step 1 · Form the Quadratic Polynomial

Given Sum of zeroes (S)=−14\text{Sum of zeroes } (S) = -\dfrac{1}{4} and Product of zeroes (P)=14\text{Product of zeroes } (P) = \dfrac{1}{4}.

The general form of the quadratic polynomial is

P(x)=x2−(sum)x+(product)=x2−(−14)x+(14)=x2+14x+14\begin{aligned} P(x) &= x^2 - (\text{sum})x + (\text{product}) \\[0.6em] &= x^2 - \left(-\dfrac{1}{4}\right)x + \left(\dfrac{1}{4}\right) \\[0.6em] &= x^2 + \dfrac{1}{4}x + \dfrac{1}{4} \end{aligned}

Multiplying by 44 to clear the denominator

P(x)=4(x2+14x+14)=4x2+x+1\begin{aligned} P(x) &= 4\left(x^2 + \dfrac{1}{4}x + \dfrac{1}{4}\right) \\[0.6em] &= 4x^2 + x + 1 \end{aligned}
Answer

(v) 4x2+x+14x^2 + x + 1

(vi) 4,14, 1

Step 1 · Form the Quadratic Polynomial

Given Sum of zeroes (S)=4\text{Sum of zeroes } (S) = 4 and Product of zeroes (P)=1\text{Product of zeroes } (P) = 1.

The general form of the quadratic polynomial is

P(x)=x2−(sum)x+(product)=x2−(4)x+(1)=x2−4x+1\begin{aligned} P(x) &= x^2 - (\text{sum})x + (\text{product}) \\[0.6em] &= x^2 - (4)x + (1) \\[0.6em] &= x^2 - 4x + 1 \end{aligned}
Answer

(vi) x2−4x+1x^2 - 4x + 1

Common Mistakes
  • Sign Error in Formula: Writing x2+(sum)x+(product)x^2 + (\text{sum})x + (\text{product}) instead of the correct formula x2−(sum)x+(product)x^2 - (\text{sum})x + (\text{product}).
  • Zeroes vs. Sum/Product: Mistaking the given numbers for the individual zeroes (α\alpha and β\beta) instead of their sum (α+β\alpha + \beta) and product (αβ\alpha\beta).
  • Handling Negative Sums: In part (v), failing to account for the double negative: −(−1/4)x=+14x-(-1/4)x = +\frac{1}{4}x.

More questions in Exercise 2.2

Q1

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) x2−2x−8x^2 - 2x - 8

(ii) 4s2−4s+14s^2 - 4s + 1

(iii) 6x2−3−7x6x^2 - 3 - 7x

(iv) 4u2+8u4u^2 + 8u

(v) t2−15t^2 - 15

(vi) 3x2−x−43x^2 - x - 4

Q2

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i) 14,−1\dfrac{1}{4}, -1

(ii) 2,13\sqrt{2}, \dfrac{1}{3}

(iii) 0,50, \sqrt{5}

(iv) 1,11, 1

(v) −14,14-\dfrac{1}{4}, \dfrac{1}{4}

(vi) 4,14, 1

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