Quadratic Equations:
Introduction:
Quadratic equation is a polynomial equation of the 2nd degree. The word ‘quadratic’ comes from the Latin word quadratus which represent the word ‘square’. Quadratic equation is an equation which can be written in the form of ax2 + bx + c = 0, where a and b are coefficients of x2 and x respectively and c is a constant. The coefficient of x2 cannot be zero and the expression ax2 + bx + c is called as ‘quadratic expression’. For example, 4x2 + 5x + 2 = 0 is a quadratic equation while 5x + 2 is not a quadratic equation To find the roots of the equation we’ve to use the formula, (-b±√(b^2-4ac))/2a Sum of the root = -b/a and Product of the root = c/a, where a, b & c are real numbers.
Nature of the roots:…show more content… if the speed of the slow train is 30 km/h less than that of the fast train, then what are the speed of the two trains (in km/h)? 90, 60 b) 80, 50 c) 70, 40 d) 60, 70
Answer:
Let the speed of the faster train = x km/h and
The speed of the slower train = (x – 30) km/h
Time taken by the faster train ~ time taken by the slower train = 5 hours
i.e., 900/(x-30)- 900/x=5 ⇒ x(x – 30) = 5400 x2 – 30x – 5400 = 0 ⇒ (x + 60)(x – 90) = 0
Either x = -60 or x = 90, x cannot be negative
Thus, the speeds are 90 km/h and 60 km/h So, the answer is (a) Find the maximum value of 3 – 6x – 8x2 and the corresponding value of x, if x be real. 33/8, 3/8 b) -33/8, 3/8 c) -33/8, -3/8 d) 33/8, -3/8
Answer:
Let us take y = 3 – 6x – 8x2 ⇒ 8x2 + 6x + y – 3 = 0 If x is real, then the discriminant b2 – 4ac ≥ 0 i.e., 62 – 4*8*(y – 3) ≥…show more content… Then x=√(6+x) x2 = 6 + x ⇒ x2 –x – 6 = 0 ⇒ (x + 2) (x – 3) = 0 x = -2 or x = 3, but x cannot be negative. So, x = 3
So, the answer is (c) In a school there are two teachers. While solving the quadratic equation in p, one of them copied the constant term incorrectly and got the roots 3 and 2. The other copied the constant term and the coefficient of p2 correctly as -6 and 1 respectively. Find the correct roots. 3, -2 b) 6, -1 c) -3, 2 d) -6, 1
Answer:
Let a and b be the two roots of the equation. Then a + b = 5 and a*b = -6 The form of the quadratic equation is p2 – (a + b) p + ab = 0 p2 – 5p – 6 = 0 ⇒ (p – 6) (p + 1) = 0 ⇒ p = 6 or p = -1 The roots are 6 and -1
So, the answer is (b) Find the value of r if the product of the roots of the equation rx2 + 6x + (2r – 1) = 0 1 b) -1 c) 1/3 d) -1/3
Answer:
Product = c/a = (2r – 1)/r = -1 (given) 2r – 1 = -r ⇒ 3r = 1 ⇒ r = 1/3
So, the answer is (c)