Create a quadratic polynomial, whose sum and product of zeroes are 1 and 3/2 respectively.

Step 1: Create a quadratic polynomial (Question 13). A quadratic polynomial with sum of zeroes S and product of zeroes P is given by the formula x^2 - Sx + P. Given sum of zeroes S = 1. Given product of zeroes P = (3)/(2). Substitute these values into the formula: x^2 - (1)x + (3)/(2) To remove the fraction, we can multiply the entire polynomial by 2: 2(x^2 - x + (3)/(2)) = 2x^2 - 2x + 3 The quadratic polynomial is: 2x^2 - 2x + 3 Step 2: Analyse the nature of roots (Question 14). The given quadratic equation is 2x^2 - 5x + 1 = 0. This is in the form ax^2 + bx + c = 0, where a=2, b=-5, and c=1. The nature of the roots is determined by the discriminant = b^2 - 4ac. Substitute the values of a, b, and c: = (-5)^2 - 4(2)(1) = 25 - 8 = 17 Since > 0, the roots are real and distinct. The nature of the roots is: Real and distinct Step 3: Define Basic Proportionality Theorem (Question 15). The Basic Proportionality Theorem (also known as Thales Theorem) states that if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides proportionally. Step 4: Find the value of the expression (Question 16). The expression is ( + )^2 + ( - )^2. Expand each term using the identities (a+b)^2 = a^2 + 2ab + b^2 and (a-b)^2 = a^2 - 2ab + b^2: ( + )^2 = ^2 + 2 + ^2 ( - )^2 = ^2 - 2 + ^2 Now, add the two expanded expressions: (^2 + 2 + ^2) + (^2 - 2 + ^2) Combine like terms: ^2 + ^2 + ^2 + ^2 + 2 - 2 2^2 + 2^2 Factor out 2: 2(^2 + ^2) Using the trigonometric identity ^2 + ^2 = 1: 2(1) = 2 The value of the expression is: 2 Step 5: Define angle of elevation (Question 17). The angle of elevation is the angle formed by the line of sight with the horizontal when an observer looks upwards at an object. Step 6: Find the radius of the circle (Question 18). Let O be the center of the circle, A be the external point, and T be the point of tangency on the circle. The length of the tangent from point A to the circle is AT = 4 cm. The distance from point A to the center of the circle is OA = 5 cm. The radius of the circle is OT. We know that the radius is perpendicular to the tangent at the point of tangency, so OTA is a right-angled triangle with the right angle at T. By the Pythagorean theorem: OA^2 = OT^2 + AT^2 Substitute the given values: (5\, cm)^2 = OT^2 + (4\, cm)^2 25\, cm^2 = OT^2 + 16\, cm^2 Subtract 16\, cm^2 from both sides: OT^2 = 25\, cm^2 - 16\, cm^2 OT^2 = 9\, cm^2 Take the square root of both sides: OT = sqrt(9\, cm)^2 OT = 3\, cm The radius of the circle is: 3\, cm