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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)

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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

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(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:

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Squaring on both sides
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only feasible solution is x = 4n

Question 4.Solve: (2 sqrtfracxa+3 sqrtfracax=fracba+frac6 ab)Solution:

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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,

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(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.
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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.

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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

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Question 4.

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Solution:
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Question 5.5x – 1 + 51 – x = 26Solution:

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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

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