For a quadratic polynomial Ax² + Bx + C, the sum of the zeroes is given by -B/A and the product of the zeroes is given by C/A. Given the polynomial x² + (a+1)x + b and zeroes 2 and 3, we can set up two equations. The sum of zeroes is 2+3=5, which equals -(a+1). The product of zeroes is 2 × 3=6, which equals b. Solving these equations gives a=-6 and b=6.

For a quadratic polynomial Ax² + Bx + C, the sum of the zeroes is given by -B/A and the product of the zeroes is given by C/A. Given the polynomial x² + (a+1)x + b and zeroes 2 and 3, we can set up two equations. The sum of zeroes is 2+3=5, which equals -(a+1). The product of zeroes is 2 × 3=6, which equals b. Solving these equations gives a=-6 and b=6.

For the polynomial x² + px + 45, let the zeroes be α and β. The sum of zeroes is α + β = -p and the product of zeroes is α β = 45. We are given that the square of the difference of the zeroes is 144, i.e., (α - β)² = 144. We use the identity (α - β)² = (α + β)² - 4αβ. Substituting the values, we get (-p)² - 4(45) = 144, which simplifies to p² - 180 = 144. Solving for p, we find p² = 324, so p = ± 18.

For the polynomial x² + px + 45, let the zeroes be α and β. The sum of zeroes is α + β = -p and the product of zeroes is α β = 45. We are given that the square of the difference of the zeroes is 144, i.e., (α - β)² = 144. We use the identity (α - β)² = (α + β)² - 4αβ. Substituting the values, we get (-p)² - 4(45) = 144, which simplifies to p² - 180 = 144. Solving for p, we find p² = 324, so p = ± 18.

While playing in a garden, Samaira saw a honeycomb and asked her mother about it. Her mother replied that it is a honeycomb. Also, she told her that the shape of the honeycomb formed is a mathematical structure. The mathematical representation of the honeycomb is shown in the graph. Based on the above information, answer the following questions : (i) How many zeroes are there for the polynomial represented by the graph given ? (ii) Write the zeroes of the polynomial. (iii) (a) If the zeroes of a polynomial $x^2 + (a + 1) x + b$ are 2 and 3, then determine the values of a and b. OR (iii) (b) If the square of difference of the zeroes of the polynomial $x^2 + px + 45$ is 144, then find the value of p.

Answer: (i) There are 3 zeroes. (ii) The zeroes are -2, 1, and 3. (iii) (a) $a = -6$ , $b = 6$ . (iii) (b) $p = \pm 18$ .

Step-by-step solution

Step 1: Identify Zeroes from Graph

The zeroes of a polynomial are the $x$ -values where the graph intersects the $x$ -axis. By observing the given graph, we can see that the curve crosses the $x$ -axis at three distinct points: $x = -2$ , $x = 1$ , and $x = 3$ . Therefore, there are three zeroes for the polynomial.

Step 2: Solve Part (iii) (a): Find a and b using sum and product of zeroes

For a quadratic polynomial $Ax^2 + Bx + C$ , the sum of the zeroes is given by $-B/A$ and the product of the zeroes is given by $C/A$ . Given the polynomial $x^2 + (a+1)x + b$ and zeroes 2 and 3, we can set up two equations. The sum of zeroes is $2+3=5$ , which equals $-(a+1)$ . The product of zeroes is $2 \times 3=6$ , which equals $b$ . Solving these equations gives $a=-6$ and $b=6$ .

Step 3: Solve Part (iii) (b): Find p using difference of zeroes

For the polynomial $x^2 + px + 45$ , let the zeroes be $\alpha$ and $\beta$ . The sum of zeroes is $\alpha + \beta = -p$ and the product of zeroes is $\alpha \beta = 45$ . We are given that the square of the difference of the zeroes is 144, i.e., $(\alpha - \beta)^2 = 144$ . We use the identity $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$ . Substituting the values, we get $(-p)^2 - 4(45) = 144$ , which simplifies to $p^2 - 180 = 144$ . Solving for $p$ , we find $p^2 = 324$ , so $p = \pm 18$ .

More from this chapter

All Polynomials solutions →

Solve your own maths question free →