Samacheer Class 12 Maths - Theory of Equations

Let the cube side be a (>0). New cuboid sides: a+1,a+2,a+3. Increase in volume: (a+1)(a+2)(a+3)-a^3 = 52. Expand: a^3+6a^2+11a+6 - a^3 = 6a^2+11a+6 = 52. So 6a^2+11a-46=0. Discriminant Δ = 11^2+4·6·46 = 121+1104=1225=(35)^2. Thus a = (-11±35)/(12) gives a=2 or a=-23/6. Reject negative. So a=2 and cuboid volume = (3)(4)(5)=60.

(i) (x-1)(x-2)(x-3)=x^3-6x^2+11x-6. (ii) (x-1)^2(x+2)= (x^2-2x+1)(x+2)=x^3-3x+2. (iii) (x-2)(x-1/2)(x-1)= x^3-(7/2)x^2+(7/2)x-1; clearing denominators: 2x^3-7x^2+7x-2=0.

Assume original is monic x^3+2x^2+2x+3=0 so S1 = α+β+γ = -2, S2=αβ+βγ+γα=2, S3=αβγ=-3. (i) For roots 2α etc: sum =2S1=-4, pair sum=4S2=8, product=8S3=-24. Monic polynomial: x^3 - (sum)x^2 + (pair)x - (product) = x^3+4x^2+8x+24=0. (ii) For reciprocals: sum = S2/S3 = 2/(-3)=-2/3, pair sum = S1/S3 = (-2)/(-3)=2/3, product = 1/S3 = -1/3. Multiply by 3: 3x^3+2x^2+2x+1=0. (iii) For negatives: sum = -S1=2, pair sum=S2=2, product = -S3=3 giving x^3-2x^2+2x-3=0.

For 3x^3-16x^2+23x-6=0, product of all roots = -d/a = -(-6)/3 = 2. If two roots have product 1, the third root = 2/1 =2, so x=2 is a root. Divide polynomial by (x-2): quotient 3x^2-10x+3. Solve 3x^2-10x+3=0 ⇒ x=(10±√(100-36))/6=(10±8)/6 → x=3 or x=1/3. So roots 2,3,1/3 (two of which multiply to 1: 3·1/3=1).

For 2x^3-8x^2+6x-3=0, sum of roots S1 = -b/a = 8/2 =4; sum of pairwise S2 = c/a = 6/2 =3. Sum of squares = S1^2 - 2S2 = 4^2 - 2·3 = 16 - 6 = 10.

Let the two roots in ratio 3:2 be 3t and 2t, and third root be u. For a monic cubic x^3+ax^2+bx+c=0 we have: 3t+2t+u = -a, 6t^2+5tu = b, 6t^2 u = -c. From the given polynomial (assumed) compare coefficients and solve these three equations for t and u, then factor. (Numerical reconstruction of the original polynomial is ambiguous from OCR; please confirm the exact equation for a numeric solution.)

Σ α/(βγ) = α/(βγ)+β/(γα)+γ/(αβ) = (α^2+β^2+γ^2)/(αβγ). Now α^2+β^2+γ^2 = (α+β+γ)^2 -2(αβ+βγ+γα) = (-b/a)^2 -2(c/a) = (b^2 -2ac)/a^2. Also αβγ = -d/a. Hence Σ α/(βγ) = [(b^2-2ac)/a^2] / (-d/a) = (b^2-2ac)/(-ad) = (2ac-b^2)/(ad).

If the quartic has roots α,β,γ,δ then the requested quadratic with roots r1 = S1 and r2 = P is t^2 - (r1+r2)t + r1 r2 = 0, i.e. t^2 - (S1+P)t + S1·P = 0. Replace S1 and P by expressions in the quartic's coefficients (S1 = - (coeff of x^3)/(coeff of x^4), P = constant/(coeff of x^4) times appropriate sign). (The OCRed quartic was unclear; substitute the actual coefficients to obtain integer coefficients.)

If r is common root then r^2 + pr + q = 0 and r^2 + p'r + q' = 0. Subtract: (p-p')r + (q-q') = 0 ⇒ r = (q' - q)/(p - p') = (pq' - p'q)/(q' - q) (by eliminating r^2 and rearranging; both forms equivalent algebraically). For the tree: let standing part = x, cut part = 12-x and x = ∛(12-x). Cubing: x^3 = 12 - x ⇒ x^3 + x - 12 = 0; solve for real positive root.

For ax^2+bx+c with a=2,b=k,c=k, discriminant Δ = k^2 - 4·2·k = k^2 -8k = k(k-8). So (i) Δ>0 ⇒ k<0 or k>8: two distinct real roots. (ii) Δ=0 ⇒ k=0 or k=8: equal real roots. (iii) 0<k<8 ⇒ Δ<0: two complex conjugate roots.

Complex conjugate 2-3i is also a root. Minimal polynomial is (x-(2+3i))(x-(2-3i)) = (x-2)^2 + 9 = x^2 -4x +13.

Same as previous: include the conjugate 2-3i; minimal polynomial (x-(2+3i))(x-(2-3i)) = x^2 -4x +13.

Let α = ∛5 -3. Then α+3 = ∛5 ⇒ (α+3)^3 =5 ⇒ expand: α^3 +9α^2 +27α +27 -5 =0 ⇒ α^3 +9α^2 +27α +22 =0. This cubic has rational coefficients and is minimal (degree 3).

Let parabola y = ax^2 + bx + c and line y = px + q. Intersection points satisfy ax^2 + bx + c = px + q ⇒ ax^2 + (b-p)x + (c-q) = 0, a quadratic in x. A quadratic has at most two distinct real roots, so the line and parabola intersect in at most two points.

If two roots are negatives of each other, say r and -r, and the third is s, then sum of roots = r-r+s = s = - (coeff x^2)/(leading coeff). Use Vieta to determine s and then use pairwise sum and product to find r; factor cubic as (x-r)(x+r)(x-s) and solve. (Exact numeric solution requires correct original polynomial; please confirm the equation.)

Let roots be a-d, a, a+d. Sum = 3a = -(-36)/9 = 4 ⇒ a = 4/3. Pairwise sum = (a-d)a + a(a+d) + (a-d)(a+d) = 3a^2 - d^2 = c/a = 44/9. Substitute a to find d^2: 3(16/9)-d^2 =44/9 ⇒ 48/9 - d^2 =44/9 ⇒ d^2 =4/9 ⇒ d = ±2/3. Thus roots are a-d=4/3 -2/3 =2/3, a=4/3, a+d=2. So roots: 2/3, 4/3, 2. (Verify these satisfy original polynomial.)

Let roots be ar^2, ar, a (in G.P.). Sum = a(r^2+r+1) = -(-26)/3 = 26/3. Pairwise sum = a^2(r^3 + r^2 + r) ? Use Vieta to get equations for a and r and solve. (Exact numeric solution needs the precise polynomial; please confirm the equation from the book.)

Let roots be α,β,γ with α = 2(β+γ). Using α+β+γ = -b/a and other Vieta relations one gets equations to determine k and the roots. Substitute α in terms of β+γ and solve the resulting system. (Exact numeric final requires the precise given polynomial; please confirm.)

Given 1+2i is a root, so is 1-2i. Also 3 is a root. Multiply corresponding linear factors (x-3)[(x-1)^2+4] = (x-3)(x^2-2x+5). Divide the given polynomial by this cubic to get a cubic quotient; factor the quotient to obtain remaining zeros. (Because the OCR of the polynomial is unclear, I cannot complete the numeric division here.)

(i) For 2x^3-9x^2+10x-3, try rational roots ±1,±3,±1/2,±3/2. Substitute x=1: 2-9+10-3=0, so x=1 is a root. Divide by (x-1): quotient 2x^2-7x+3 ⇒ roots x=(7±√(49-24))/4=(7±5)/4 ⇒ x=3 or x=1/2. So roots 1,3,1/2. (ii) For 8x^3-2x^2-7x+3 try rational roots ±1,±3,±1/2,±3/2,±1/4,±3/4. Test x=1:8-2-7+3=2 ≠0; x=1/2:1 -0.5 -3.5 +3 =0 ⇒ x=1/2 root. Divide polynomial by (x-1/2): quotient 8x^2+2x-6 => divide by 2: 4x^2+x-3=0 ⇒ x = [-1±√(1+48)]/8 = (-1±7)/8 ⇒ x = 3/4 or x = -1. So roots: 1/2, 3/4, -1.

The provided text is not unambiguous. Please resend the equation in clear form (e.g. x^4 - 2x^3 + 14x^2 - 45x + 36 = 0) so it can be solved.

(i) Pair factors: (x-5)(x+4)=x^2 - x -20, (x-7)(x+6)=x^2 - x -42. Let y = x^2 - x. Then (y-20)(y-42)=504 ⇒ y^2 -62y +336 =0. Discriminant =2500 ⇒ y = (62 ± 50)/2 ⇒ y =56 or 6. So x^2 - x -56 =0 ⇒ x = 8 or -7; x^2 - x -6 =0 ⇒ x = 3 or -2. (ii) Expanding gives x^4 -14x^3 +63x^2 -106x +56 = 16 ⇒ x^4 -14x^3 +63x^2 -106x +40 =0. This quartic has no simple integer roots; it factors into quadratics with irrational coefficients — solve by standard quartic methods or numeric methods to obtain the four roots.

The OCR'd expression cannot be reliably reconstructed. Provide the exact factorised equation (for example (x-2)(x-1)(x-3)(x+2)=20 or similar) and I will solve it.

(i) Let s = sin x. Then s^2 -5s +4 = (s-1)(s-4)=0 ⇒ s =1 or 4. Since sin x ∈ [-1,1], only s=1 is valid. Thus x = π/2 + 2kπ, k∈Z. (ii) The OCR for part (ii) is not clear; please resend the exact equation.

The OCR text is ambiguous. To apply the Rational Root Theorem we need the exact polynomial (e.g. x^3 -2x -1 =0). Please resend the precise equations.

I cannot reconstruct the intended equation from the OCR. Provide the exact expression (for example 8^x - 8^{x-n} = 63 or x^8 - 8^n = 63) and I will solve it.

The OCR'd text is not interpretable. Send the exact algebraic equation (with parentheses if needed) and I'll provide a concise solution.

The OCR output is ambiguous. Provide the exact polynomials (coefficients and powers) and I'll solve them succinctly.

The OCR must be corrected. Provide the full clear equation and I'll solve it.

Once the correct polynomial is given, use synthetic division by x - 1/3 (or multiply by 3 to get integer coefficients then divide by (3x-1)) to reduce degree and solve the remaining cubic/quadratic. Please supply the exact polynomial.

To answer, apply Descartes' Rule of Signs to the correctly written polynomial and to f(-x) to determine the maximum possible numbers of positive and negative real roots. Please provide the exact polynomial.

Provide exact polynomials for precise answer. For the guessed polynomials: (1) f(x)=x^3 - x^2 -5x +6 has sign changes: + - - + ⇒ 2 or 0 positive roots. f(-x) = -x^3 - x^2 +5x +6 ⇒ sign changes - + + ⇒ 1 negative root. Similar for the second; sketch using leading term positive (→ -∞ as x→ -∞, → +∞ as x→ +∞) and approximate turning points.

Apply Descartes' Rule of Signs to f(x) and f(-x) to bound the number of positive and negative real roots; subtract the maximum possible real roots from degree 9 to get at least 6 nonreal roots. Exact counts require the exact polynomial as printed.

Once the exact polynomial is given, use Descartes' Rule of Signs on f(x) and f(-x) to determine possible numbers of positive and negative roots.

Provide the correct polynomial; then apply Descartes' Rule and fundamental theorem of algebra to obtain counts of real and nonreal zeros.