You have the right to Download Samacheer Kalvi 12th Maths publication Solutions overview Pdf, Tamilnadu State Board aid you to review the finish Syllabus and score more marks in her examinations.

You are watching: Factor 2x^2+5x+3

## Tamilnadu Samacheer Kalvi 12th Maths options Chapter 3 concept of Equations Ex 3.5

Question 1.Solve the complying with equations:(i) sin2x – 5sinx + 4 = 0(ii) 12x3 + 8x = 29x2 – 4Solution:(i) sin² x – 5 sin x + 4 = 0Put sin x = tt² – 5t + 4 = 0(t – 1) (t – 4) = 0t = 1 or t = 4sin x = 4 or sin x = 1(is not possible) sin x = sin (fracπ2)x = nπ + (-1l)n (fracπ2) ∀ n ∈ z.

(ii) 12x3 + 8x = 29x2 – 412x3 – 29x2 + 8x + 4 = 0 ….. (1)By Trail and error method, (x – 2) is a factor of (1)The other factor is 12x2 – 5x – 2The roots is 12x2 – 5x – 2 = 0(3x – 2) (4x + 1) = 0x = (frac23), x = (-frac14)The roots space 2, (frac23), (-frac14) Question 2.Examine because that the rational root of:(i) 2x3 – x2 – 1 = 0(ii) x8 – 3x + 1 = 0Solution:(i) 2x3 – x2 – 1 = 0Sum that the co-efficients = 0∴ (x – 1) is a factorThe other element is 2x2 + x + 1.The root is (2x2 + x + 1) = 0Here ∆ = b2 – 4ac = (1)2 – 4(2) (1) = 1 – 8 = -7 The staying roots space imaginary.The just rational root is x = 1 (ii) x8 – 3x + 1 = 0 …. (1)Here one = 1, a0 = 1If (fracpq) is a rational root of (1)Then q is a element an, ns is a variable of a0The feasible values of p and q are ± 1.Among the feasible values 1, -1, <(p, q) = 1>None of castle satisfies the equation (1)The over equation has actually no rational roots.

Question 3.Solve: (8 x^frac32 n-8 x^frac-32 n=63)Solution: Squaring on both sides only feasible solution is x = 4n

Question 4.Solve: (2 sqrtfracxa+3 sqrtfracax=fracba+frac6 ab)Solution:  Question 5.Solve the equations:(i) 6x4 – 35x3 + 62x2 – 35x + 6 = 0(ii) x4 + 3x3 – 3x – 1 = 0Solution:(i) 6x4 – 35x3 + 62x2 – 35x + 6 = 0 ….. (1)It is a even degree reciprocal equation as p(x) = (x^n pleft(frac1x ight))Dividing equation (1) by x2,  (ii) x4 + 3x3 – 3x – 1 = 0 …… (1)It is one even level reciprocal function of form II.1, -1 space the systems of equation (1)(x – 1), (x + 1) space the element of (1)(x2 – 1) is a element of (1)Dividing (1) by (x2 – 1)we get, x2 + 3x + 1 = 0 is the other factor.  Question 6.Find all actual numbers solve 4x – 3(2x+2) + 25 = 0Solution:4x – 3(2x+2) + 25 = 0.(2x)² – 3.(2x 2²) + 25 = 0(2x)² – 12(2x) + 32 = 0Put 2x = t(t² – 12t + 32 = 0)(y – 4)(y – 8) = 0y = 4 or y = 8t = 8 (or) t = 42x = 8 = 2³ (or) 2x = 4 = (2)²x = 3 (or) x = 2Roots are 3, 2

Question 7.Solve the equation 6x4 – 5x3 – 38x2 – 5x + 6 = 0 if that is recognized that (frac13) is a solution.Solution:6x4 – 5x3 – 38x2 – 5x + 6 = 0 …… (1)x = (frac13) is a Solution∴ (3x – 1) is a variable of (1)(1) is a reciprocal equation even degree divide (1) by x2.

See more: Is Coca Cola Good For Diarrhea ? Is It Ok To Drink Coke When You Have Diarrhea  ### Samacheer Kalvi 12th Maths remedies Chapter 3 theory of Equations Ex 3.5 extr Problems

Question 1.Solve: 32x + 4 + 1 = 2.3x + 2Solution:32x + 4 = 3x + 2 + 3x + 2 – 132x + 4 – 3x + 2 = 3x + 2 – 1 ⇒ 3x + 2<3x + 2 – 1> = <3x + 2 – 1>3x + 2= 1 ⇒ 3x + 2 = 30x + 2 = 0 ⇒ x = -2

Question 2.Solve: 2x – 2x + 3 + 24 = 0.Solution:22x – (2x.23) + 24 = 0 because 23 = 8 = 4 + 4 = 22 + 2222x – (4 + 4)2x + 24 = 0 ⇒ 22x – (22 + 22)2x + 24 = 0(2x – 22)(2x – 22) = 0 ⇒ (2x – 22)2 = 02x = 22 ⇒ x = 2

Question 3.Solve: (x – 4) (x + 2) (x + 3) (x – 3) + 8 = 0.Solution:(x – 4) (x + 3) (x + 2) (x – 3) + 8 = 0(x2 – x – 12) (x2 – x – 6) + 8 = 0Let y = x2 – x(y – 12) (y – 6) + 8 = 0 ⇒ y2 – 18y + 72 + 8 = 0y2 – 18y + 80 = 0 ⇒ (y – 10)( y – 8) = 0Case (i) y – 10 = 0x2 – x – 8 = 0 Question 4. Solution:  Question 5.5x – 1 + 51 – x = 26Solution: Question 6.Solve: 12x4 – 56x3 + 89x2 – 56x + 12 = 0Solution:Since the coefficients that the equations room equal native both ends.Divide the equation by x2  