25 ++ p(x)=3x solution 372150-P(x)=3x solution
Solution set = {15 ≤ x < 5} The solution set can be graphed on the number line as Question 12 Solve and graph the solution set of (i) 2x – 9 < 7 and 3x 9 ≤ 25, x ∈ R (ii) 2x – 9 ≤ 7 and 3x 9 > 25, x ∈ I (iii) x 5 ≥ 4(x – 1) and 3 – 2x < 7, x ∈ R Solution Question 13 Solve and graph the solution set of (i) 3x(iv) p(x)=kx 2 –3xk Solution If x1 is a factor of p(x), then p(1) = 0 By Factor Theorem ⇒ k(1) 2 –3(1)k = 0 ⇒ k−3k = 0 ⇒ 2k−3 = 0 ⇒ k= 3/2 4 Factorize (i) 12x 2 –7x1 Solution Using the splitting the middle term method, We have to find a number whose sum = 7 and product =1×12 = 12If p(x) is a third degree polynomial and q(x) is a fourth degree polynomial then find the degree of p(x) × q(x) Answer i p(x) = x 2 3x 1 , q(x) = 2x – 4
If P X X3 3x2 5x 3 And G X X2 2 Find Remainder Mathematics Topperlearning Com T7tfrb44
P(x)=3x solution
P(x)=3x solution-If P is the solution set of – 3x 4 < 2x – 3, x ∈ N and Q is the solution set of 4x – 5 < 12, x ∈ W, find (i) P ∩ Q (ii) Q – P Answer 33 (i) – 3 x 4 < 2 x – 3 – 3x – 2x < – 3 – 4 => – 5x < – 7 Question 34 A = {x 11x – 5 > 7x 3, x ∈R} and B = {x 18x – 9 ≥ 15 12x, x ∈R}Is x = 5 a solution of the quadratic equation x 2 – 2x – 15 = 0?
Remember that if f(x) = Ax^n, then f ' (x) = nAx^(n1) With this in mind, we have P(x) = 3x^2 5x^3 7x^4 3x^5 So P ' (x) = 6x 15x^2 28x^3 15x^4Simplifying p (x) = 3x 14 Multiply p * x px = 3x 14 Reorder the terms px = 14 3x Solving px = 14 3x Solving for variable 'p' Move all terms containing p to the left, all other terms to the right Divide each side by 'x' p = 14x 1 3 Simplifying p = 14x 1 3 Reorder the terms p = 3 14x 1Swap sides so that all variable terms are on the left hand side 3x1=y 3x − 1 = y Add 1 to both sides Add 1 to both sides 3x=y1 3x = y 1 Divide both sides by 3 Divide both sides by 3
Ie say the root is x,then x>=0 As per the question,one of the three roots is 2/3 if all three roots are real,then (2/3) cannot be one of the roots Please check your question and post againPrabhat Edit The three roots (whether real or not) can be found by equating p(x) to zero and then solving the trinomialYou can solve as many equations as you like completely free If you need detailed stepbystep answers you'll have to sign up for Mathway's premium service (provided by a third party)Here, we are given three terms, First term (a 1) = 2xSecond term (a 2) = x 10 Third term (a 3) = 3x 2We need to find the value of x for which these terms are in APSo, in an AP the difference of two adjacent terms is always constant
Find all real zeros of P(x) If neither of these were the instructions, post again and this time give the instructions as to what you are supposed to do with P(x) P(x) = x 5 3x 3 2x 6 x 5 and 3x 3 are both positive, so there is no sign change going from x 5 to 3x 3 3x 3 and 2x are both positive, so there is no sign change going from 3x 3 to 2x 2x is positive and 6 is negative, so that is 1 sign change going from 2x to 6 So there is exactly 1 positive zero of P(x)Solution 2(1) t> Solution 2(2) Solution 2(3) Solution 3 If we divide the total number of chocolates that the student had (x 4 – 3x 3 5x 2 8x 5) by the number of chocolates received by each friend (x 2 – 1), the quotient will give the number of friends and the remainder will give the number of chocolates left over for his teacher Thus, the number of friends = x 2 – 3x 6If p(x)= x 2 – 2√2x1, then find the value of p(2√2) Solution Put x = 2√2 in p(x), we obtain p(2√2) = (2√2) 2 – 2√2(2√2) 1 = (2√2) 2 – (2√2) 2 1 = 1 Question 4 Find the value of m, if x 4 is a factor of the polynomial x 2 3x m Solution Let p(x) = x 2 3x m Since (x 4) or (x – (4)} is a factor of p
All the solutions of Polynomials Mathematics explained in detail by experts to help students prepare for their CBSE exams R S Aggarwal and V Aggarwal Solutions for Class 9 Mathematics CBSE, 2 Polynomials When p(x) = x 3 3x 2 4x 32 is divided by (x 2), the remainder is (a) 0 (b) 32 (c) 36 (d) 4 Solution 18StepbyStep Solutions Use stepbystep calculators for chemistry, calculus, algebra, trigonometry, equation solving, basic math and more Gain more understanding of your homework with steps and hints guiding you from problems to answers!X^45x^24=0 \sqrt{x1}x=7 \left3x1\right=4 \log _2(x1)=\log _3(27) 3^x=9^{x5} equationcalculator 3x5=14 en Related Symbolab blog posts High School Math Solutions – Quadratic Equations Calculator, Part 1 A quadratic equation is a second degree polynomial having the general form ax^2 bx c = 0, where a
If p(x) is a third degree polynomial and q(x) is a fourth degree polynomial then find the degree of p(x) × q(x) Answer i p(x) = x 2 3x 1 , q(x) = 2x – 4Here, P=y 2 ,Q=xy , R= x(z2y) Hence the general solution is f(x 2 y 2, yzy 2)=0 3Solve(3z4y)p(4x2z)q=2y3x Soln Given (3z4y)p(4x2z)q=2y3x This equation of the form Ppq=R Here, P= (3z4y) ,Q=(4x2z) , R= 2y3x Again use Lagrangian multipliers 2,3,4, 2x3y4z=b Hence the general solution is, F(x 2 y 2 z 2 , 2x3y4z)=0Solution Let p (x) = 3x 3 11x 2 34x 106 p (x) in standard form Coefficients are 3 11 34 106 q(x) = x – 3, its zero x = 3 Synthetic division Quotient is 3x 2 x 94, it is compared with the given quotient 3x 2 ax b Coefficient of x is a = Constant term is b = 94 Remainder r = 3
The remainder is zero when f(x) is divided by (x – a) f(a) = 0 The solution to f(x) = 0 is a The zero of the function f(x) is a Example Determine whether x 1 is a factor of the following polynomials a) 3x 4 x 3 – x 2 3x 2 b) x 6 2x(x – 1) – 4 Solution a) Let f(x) = 3x 4 x 3 – x 2 3x 2 f(–1) = 3(–1)4 (–1Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, historySolution x 2 – 2x – 15 = 0 For x = 5 to be solution of the given quadratic equation it should satisfy the equation So, substituting x = 5 in the given equation, we get LHS = (5) 2 – 2(5) – 15 = 25 – 10 – 15 = 0 = RHS
3 Find the indicated derivative for each function Show your work (Sec 26) Find y000, y= 4x4 x2 7 Solution y0= 16x3 2x y00= 48x2 2 y000= 96x Find d2f dx2, where f(x) = xsinx Solution f0(x) = xcosx sinx f00(x) = cosx xsinx cosx= 2cosx xsinx Find r(4), where r(x) = 1=x Solution r0(x) = 1=x2 r00(x) = 2=x3 r000(x) = 6=x4 r(4)(x) = 24=x5 4 Find a tangent line through the curve 2x 2Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with stepbystep explanations, just like a math tutorSolve any equation with this free calculator!
Step by step solution Step 1 Equation at the end of step 1 3x • (x 2) = 0 Step 2 Theory Roots of a product 21 A product of several terms equals zero When a product of two or more terms equals zero, then at least one of the terms must be zero We shall now solve each term = 0 separatelyFind g(x) Solution Divided, p(x) = x 3 – 3x 2 x 2 Quotient = (x – 2) Remainder = (2x 4) Let g(x) be the divisor According to the division algorithm, Dividend = Divisor x Quotient Remainder Concept Insight When a polynomial is divided by any other nonzero polynomial, then it satisfies the division algorithm which is as belowIs x = 5 a solution of the quadratic equation x 2 – 2x – 15 = 0?
A) no solution B) one solution C) two solutions D) infinitely many solutions 9) Which graph represents the solution for the equation 1 2 x 1 = x 2?If P is the solution set of – 3x 4 < 2x – 3, x ∈ N and Q is the solution set of 4x – 5 < 12, x ∈ W, find (i) P ∩ Q (ii) Q – P Solution (i) – 3 x 4 < 2 x – 3 – 3x – 2x < – 3 – 4 => – 5x < – 7 Question 34 A = {x 11x – 5 > 7x 3, x ∈R} and B = {x 18x – 9 ≥ 15 12x, x ∈R}Solution (D) Let p(x) = ax 2 bx c be the required polynomial whose zeroes are 2 and 5 From (i) and (ii), a = 1,b = 3 and c = 10 ∴p(x) = x 2 – 3x – 10 But we know that, if we multiply/divide any polynomial by any arbitrary constant, then the zeroes of polynomial never change ∴ p(x) = kx 2 – 3kx – 10k, where, k is a real number
Here we will look at solving a special class of Differential Equations called First Order Linear Differential Equations First Order They are "First Order" when there is only dy dx, not d 2 y dx 2 or d 3 y dx 3 etc Linear A first order differential equation is linear when it can be made to look like this dy dx P(x)y = Q(x) Where P(x) and Q(x) are functions of x To solve it there is aIf α, β Are the Zeros of the Polynomial P(X) = 4x2 3x 7, Then 1 α 1 β is Equal to CBSE CBSE Class 10 Question Papers 6 Textbook Solutions Important Solutions 3114 CBSE Previous Year Question Paper With Solution for Class 12 Commerce;Hence, p(0)= 1, p(1) = 0, p(2) = 3, for the equation p(x) = (x−1)(x1) Question 3 Verify whether the following are zeroes of the polynomial, indicated against them (i) p(x) = 3x1, x=−1/3
A) A B) B C) C D) D * Hover over answer image to enlarge 10) x 3y = 5x 6y = 4Remember that if f(x) = Ax^n, then f ' (x) = nAx^(n1) With this in mind, we have P(x) = 3x^2 5x^3 7x^4 3x^5 So P ' (x) = 6x 15x^2 28x^3 15x^4The remainder is zero when f(x) is divided by (x – a) f(a) = 0 The solution to f(x) = 0 is a The zero of the function f(x) is a Example Determine whether x 1 is a factor of the following polynomials a) 3x 4 x 3 – x 2 3x 2 b) x 6 2x(x – 1) – 4 Solution a) Let f(x) = 3x 4 x 3 – x 2 3x 2 f(–1) = 3(–1)4 (–1
Solution The degree of zero polynomial is not defined Question 8 Find the value of the polynomial P(x) = 3x 4 5x 3 5x 2 7x 3 at x = 1 33 55 Correct Option is 2 Solution Given, polynomial is P(x) = 3x 4 5x 3 5x 2 7x 3 Now, at x = 1 P(1) = 3*1 4 5*1 3 5*1 2 7*1 3 = 3 5 5 7 3 = 10 13 = 3Example Divide 3x 3 – 8x 5 by x – 1 Solution The Dividend is 3x 3 – 8x 5 and the divisor is x – 1 After this, the leading term of the dividend is divided by the leading term of the divisor ie 3x 3 ÷ x =3x 2 This result is multiplied by the divisor ie 3x 2 (x 1) = 3x 33x 2 and it is subtracted from the divisor Now again, this result is treated as a dividend and theHow many solutions can be found for the system of linear equations represented on the graph?
Solution for P(X)=3X^2X4 equation Simplifying P(X) = 3X 2 X 4 Multiply P * X PX = 3X 2 X 4 Reorder the terms PX = 4 X 3X 2 Solving PX = 4 X 3X 2 Solving for variable 'P' Move all terms containing P to the left, all other terms to the right Divide each side by 'X' P = 4X1 1 3X Simplifying P = 4X1 1 3X Reorder the terms P = 1 4X1 3XHere we will look at solving a special class of Differential Equations called First Order Linear Differential Equations First Order They are "First Order" when there is only dy dx, not d 2 y dx 2 or d 3 y dx 3 etc Linear A first order differential equation is linear when it can be made to look like this dy dx P(x)y = Q(x) Where P(x) and Q(x) are functions of x To solve it there is aAfter you enter the expression, Algebra Calculator will plug x=6 in for the equation 2x3=15 2(6)3 = 15 The calculator prints "True" to let you know that the answer is right More Examples Here are more examples of how to check your answers with Algebra Calculator Feel free to try them now For x6=2x3, check (correct) solution x=3 x6=2x
(ii) Find 'a' if the two polynomials ax 3 3x 2 – 9 and 2x 3 4x a, leaves the same remainder when divided by x 3 Solution Question 9 By factor theorem, show that (x 3) and (2x – 1) are factors of 2x 2 5x – 3 Solution Question 10 Show that (x – 2) is a factor of 3x 2 – x – 10 Hence factorise 3x 2 – x – 10Ie say the root is x,then x>=0 As per the question,one of the three roots is 2/3 if all three roots are real,then (2/3) cannot be one of the roots Please check your question and post againPrabhat Edit The three roots (whether real or not) can be found by equating p(x) to zero and then solving the trinomialMathematics 6 Solutions for HWK 22b Section 84 p399 Problem 1, §84 p399 Let T P 2 −→ P 3 be the linear transformation defined by T(p(x)) = xp(x) (a) Find the matrix for T with respect to the standard bases
If the roots of the equation x 2 p x c = 0 are (2, − 2) and the roots of the equation x 2 b x q = 0 are (− 1, − 2), then the roots of the equation x 2 b x c = 0 are Medium View solutionWithout solving the following quadratic equation, find the value of 'p' for which the given equation has real and equal roots x 2 (p 3)x p = 0 Asked by Topperlearning User 27th Oct, 17, 0224 PMSolution x 2 – 2x – 15 = 0 For x = 5 to be solution of the given quadratic equation it should satisfy the equation So, substituting x = 5 in the given equation, we get LHS = (5) 2 – 2(5) – 15 = 25 – 10 – 15 = 0 = RHS
Just enter your equation carefully, like shown in the examples below, and then click the blue arrow to get the result!If the roots of the equation x 2 p x c = 0 are (2, − 2) and the roots of the equation x 2 b x q = 0 are (− 1, − 2), then the roots of the equation x 2 b x c = 0 are Medium View solutionIf P is the solution set of – 3x 4 < 2x – 3, x ∈ N and Q is the solution set of 4x – 5 < 12, x ∈ W, find (i) P ∩ Q (ii) Q – P Answer 33 (i) – 3 x 4 < 2 x – 3 – 3x – 2x < – 3 – 4 => – 5x < – 7 Question 34 A = {x 11x – 5 > 7x 3, x ∈R} and B = {x 18x – 9 ≥ 15 12x, x ∈R}
Let P (x) = (x − 3) (x − 5) Q (x) x A B P (3) = 1 0 = 3 A B P (5) = 6 = 5 A B This gives A = − 2, B = 1 6 Thus by division by (x − 5) (x − 3), ans will be − 2 x 1 6 More Items ShareHere, P=y 2 ,Q=xy , R= x(z2y) Hence the general solution is f(x 2 y 2, yzy 2)=0 3Solve(3z4y)p(4x2z)q=2y3x Soln Given (3z4y)p(4x2z)q=2y3x This equation of the form Ppq=R Here, P= (3z4y) ,Q=(4x2z) , R= 2y3x Again use Lagrangian multipliers 2,3,4, 2x3y4z=b Hence the general solution is, F(x 2 y 2 z 2 , 2x3y4z)=0
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