To find the cubic equation whose roots are the squares of the roots of the given equation \( 12x^3 - 20x^2 + x + 3 = 0 \), we proceed as follows: Step 1: Let the roots of the original equation Let the roots of the original equation \( 12x^3 - 20x^2 + x + 3 = 0 \) be \( r, s, t \). Then: \[ r + s + t = \frac{20}{12} = \frac{5}{3}, \quad rs + rt + st = \frac{1}{12}, \quad rst = -\frac{3}{12} = -\frac{1}{4}. \] Step 2: Find the roots of the new equation The roots of the new equation are the squares of the roots of the original equation, i.e., \( r^2, s^2, t^2 \). We need to find the sums and products of these new roots. #Sum of the new roots: \[ r^2 + s^2 + t^2 = (r + s + t)^2 - 2(rs + rt + st) = \left(\frac{5}{3}\right)^2 - 2\left(\frac{1}{12}\right) = \frac{25}{9} - \frac{1}{6} = \frac{50}{18} - \frac{3}{18} = \frac{47}{18}. \] #Sum of the products of the new roots: \[ r^2s^2 + r^2t^2 + s^2t^2 = (rs + rt + st)^2 - 2rst(r + s + t) = \left(\frac{1}{12}\right)^2 - 2\left(-\frac{1}{4}\right)\left(\frac{5}{3}\right) = \frac{1}{144} + \frac{5}{6} = \frac{1}{144} + \frac{120}{144} = \frac{121}{144}. \] #Product of the new roots: \[ r^2s^2t^2 = (rst)^2 = \left(-\frac{1}{4}\right)^2 = \frac{1}{16}. \] Step 3: Construct the new equation The new cubic equation with roots \( r^2, s^2, t^2 \) is: \[ x^3 - (r^2 + s^2 + t^2)x^2 + (r^2s^2 + r^2t^2 + s^2t^2)x - r^2s^2t^2 = 0. \] Substituting the values from Step 2: \[ x^3 - \frac{47}{18}x^2 + \frac{121}{144}x - \frac{1}{16} = 0. \] To eliminate fractions, multiply through by \( 144 \): \[ 144x^3 - 376x^2 + 121x - 9 = 0. \] Step 4: Match with the options The equation \( 144x^3 - 376x^2 + 121x - 9 = 0 \) matches option (3). Final Answer: \[ \boxed{3} \]
Solve the following pair of linear equations by the substitution method.
(i) x + y = 14
x – y = 4
(ii) s – t = 3
\(\frac{s}{3} + \frac{t}{2}\) =6
(iii) 3x – y = 3
9x – 3y = 9
(iv) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
(v)\(\sqrt2x\) + \(\sqrt3y\)=0
\(\sqrt3x\) - \(\sqrt8y\) = 0
(vi) \(\frac{3x}{2} - \frac{5y}{3}\) =-2,
\(\frac{ x}{3} + \frac{y}{2}\) = \(\frac{ 13}{6}\)
Form the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km.
(v) A fraction becomes\(\frac{ 9}{11}\), if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes \(\frac{5}{6}\). Find the fraction.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
What are X and Y respectively in the following set of reactions?
What are X and Y respectively in the following reactions?
Observe the following reactions:
The correct answer is: