If both (x – 2) and (x – 1/2) are factors of px^2 + 5x + r, then: -Maths 9th

If both (x – 2) and (x – 1/2) are factors of px^2 + 5x + r, then: -Maths 9th
by

1 Answer

Answer :

answer:
by

Related questions

If the expressions (px^3 + 3x^2 – 3) and (2x^3 – 5x + p) when divided by (x – 4) leave the same remainder, then what is the value of p ? -Maths 9th

Description : If the expressions (px^3 + 3x^2 – 3) and (2x^3 – 5x + p) when divided by (x – 4) leave the same remainder, then what is the value of p ? -Maths 9th

Last Answer : Given that the following polynomials leave the same remainder when divided by (x - 4) : We are to find the value of a. Remainder theorem: When (x - b) divides a polynomial p(x), then the remainder is p(b). So, from (i) and (ii), we get Thus, the required value of a is 1.

If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. -Maths 9th

Description : If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. -Maths 9th

Last Answer : Show that p = r.

If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. -Maths 9th

Description : If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. -Maths 9th

Last Answer : Show that p = r.

If both (x -2) and (x - 1/2) are factors of px2 + 5x + r, show that p = r. -Maths 9th

Description : If both (x -2) and (x - 1/2) are factors of px2 + 5x + r, show that p = r. -Maths 9th

Last Answer : Solution :-

If p, q, r are positive and are in A.P., the roots of quadratic equation px^2 + qx + r = 0 are real for : -Maths 9th

Description : If p, q, r are positive and are in A.P., the roots of quadratic equation px^2 + qx + r = 0 are real for : -Maths 9th

Last Answer : Given p,q,r are in A.P. then q=2p+r​.....(1). Now px2+qx+r=0 will have real root then q2−4pr≥0. or, 4(p+r)2​−4pr≥0 or, p2+r2−14pr≥0 or, r2−14rp+49p2≥48p2 or, (r−7p)2≥(43​p)2 or, (pr​−7)2≥(43​)2 [ Since p=0 for the given equation to be quadratic] or, ∣∣∣∣∣​pr​−7∣∣∣∣∣​≥43​.

If the difference in the roots of the equation x^2 – px + q = 0 is unity, then which one of the following is correct ? -Maths 9th

Description : If the difference in the roots of the equation x^2 – px + q = 0 is unity, then which one of the following is correct ? -Maths 9th

Last Answer : answer:

If x^2 + mx + n = 0 and x^2 + px + q = 0 have a common root, then the common root is -Maths 9th

Description : If x^2 + mx + n = 0 and x^2 + px + q = 0 have a common root, then the common root is -Maths 9th

Last Answer : Let α be the common root ∴α2+pα+q=0 ...........(1) and α2+qα+p=0 ........ (2) Solving (1) & (2), we get, p2−q2α2​=q−pα​=q−p1​∴α=q−pp2−q2​ and α=1 ⇒q−pp2−q2​=1 ⇒p2−q2=q−p (or) (p2−q2)+(p−q)=0 ⇒(p−q)[p+q+1]=0 ⇒p−q=0 or p+q+1=0

If the expression (px^3 + x^2 – 2x – q) is divisible by (x – 1) and (x + 1), then the values of p and q respectively are ? -Maths 9th

Description : If the expression (px^3 + x^2 – 2x – q) is divisible by (x – 1) and (x + 1), then the values of p and q respectively are ? -Maths 9th

Last Answer : Let f(x)=px3+x2−2x−q Since f(x) is divisible by (x−1) and (x+1) so x=1 and −1 must make f(x)=0. Therefore, p+1−2−q=0, i.e., p−q=1; and −p+1+2−q=0, i.e., p+q=3 Thus p=2 and q=1

If the polynomial x^6 + px^5 + qx^4 – x^2 – x – 3 is divisible by x^4 – 1, then the value of p^2 + q^2 is : -Maths 9th

Description : If the polynomial x^6 + px^5 + qx^4 – x^2 – x – 3 is divisible by x^4 – 1, then the value of p^2 + q^2 is : -Maths 9th

Last Answer : The divisor is x4−1=(x−1)(x+1)(x2+1) By factor theorem, f(1)=f(−1)=0 Thus, 1+p+q−1−1−3=0 and 1+q−1−3=p−1 i.e., p+q=4 and p−q=−2 Adding the two, 2p=2 i.e. p=1 and ∴ q=3. ∴ p2+q2=1+9=10

If (x + k) is a common factor of x^2 + px + q and x^2 + lx + m, then the value of k is -Maths 9th

Description : If (x + k) is a common factor of x^2 + px + q and x^2 + lx + m, then the value of k is -Maths 9th

Last Answer : answer:

If x^3 + px + q and x^3 + qx + p have a common factor, then which of the following is correct ? -Maths 9th

Description : If x^3 + px + q and x^3 + qx + p have a common factor, then which of the following is correct ? -Maths 9th

Last Answer : x3+px+q 3x2+p have common root Let common root =a a3+ap+q=0 a2=3−p​a=3−p​​4p3+27q2=? a3(a2+p)+q=0 ⇒3−p​​[(3−p​)+p]+q=0 3−p​​(3+2p​)=−q ⇒27−p(4p2)​=q2

If an integer P is chosen at random in the interval 0 ≤ p ≤ 5, the probability that the roots of the equation x^2 + px -Maths 9th

Description : If an integer P is chosen at random in the interval 0 ≤ p ≤ 5, the probability that the roots of the equation x^2 + px -Maths 9th

Last Answer : answer:

Two students A and B solve an equation of the form x^2 + px + q = 0. A starts with a wrong value of p and obtains the roots as 2 and 6. -Maths 9th

Description : Two students A and B solve an equation of the form x^2 + px + q = 0. A starts with a wrong value of p and obtains the roots as 2 and 6. -Maths 9th

Last Answer : Let αα and ββ be the roots of the quadratic equation x2+px+q=0x2+px+q=0 Given that, A starts with a wrong value of p and obtains the roots as 2 and 6. But this time q is correct. i.e., a product of roots ... 1 Now, from Eqs. (ii) and (iii), we get α=−3 and β=−4α=−3 and β=−4 which are correct roots.

For what value of p is the coefficient of x^2 in the product (2x – 1) (x – k) (px + 1) equal to 0 and the constant term equal to 2 ? -Maths 9th

Description : For what value of p is the coefficient of x^2 in the product (2x – 1) (x – k) (px + 1) equal to 0 and the constant term equal to 2 ? -Maths 9th

Last Answer : answer:

When (x^3 – 2x^2 + px – q) is divided by (x^2 – 2x – 3), the remainder is (x – 6), What are the values of p and q respectively ? -Maths 9th

Description : When (x^3 – 2x^2 + px – q) is divided by (x^2 – 2x – 3), the remainder is (x – 6), What are the values of p and q respectively ? -Maths 9th

Last Answer : answer:

One of the factors of (25x2 – 1) + (1 + 5x)2 is -Maths 9th

Description : One of the factors of (25x2 – 1) + (1 + 5x)2 is -Maths 9th

Last Answer : (d) Now, (25x2 -1) + (1 + 5x)2 = 25x2 -1 + 1 + 25x2 + 10x [using identity, (a + b)2 = a2 + b2 + 2ab] = 50x2 + 10x = 10x (5x+ 1) Hence, one of the factor of given polynomial is 10x.

One of the factors of (25x2 – 1) + (1 + 5x)2 is -Maths 9th

Description : One of the factors of (25x2 – 1) + (1 + 5x)2 is -Maths 9th

Last Answer : (d) Now, (25x2 -1) + (1 + 5x)2 = 25x2 -1 + 1 + 25x2 + 10x [using identity, (a + b)2 = a2 + b2 + 2ab] = 50x2 + 10x = 10x (5x+ 1) Hence, one of the factor of given polynomial is 10x.

If x^3 + 5x^2 + 10k leaves remainder – 2x when divided by x^2 + 2, then what is the value of k ? -Maths 9th

Description : If x^3 + 5x^2 + 10k leaves remainder – 2x when divided by x^2 + 2, then what is the value of k ? -Maths 9th

Last Answer : x3+5x2+10k =(x2+2)(x+5)+10k−2x−10 ⇒10k−2x−10=−2x ⇒10k−10=0 or k=1.

The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Description : The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Last Answer : Given, point (2,3) lies on the line. So, the point (2, 3) is the solution of 3x - (a -1) y = 2a - 1 On putting x = 2 and y = 3 in given solution. ∴ 3 2 - (a-1) 3 = 2a - 1 ⇒ 6 - 3a + 3 = 2a - 1 ⇒ - 3a ... 2 2) 3 = 3b ⇒ 10 - 9 = 3b ⇒ 1 = 3b ⇒ 1 / 3 = b Hence, the value of b is 1 / 3.

The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Description : The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Last Answer : Given, point (2,3) lies on the line. So, the point (2, 3) is the solution of 3x - (a -1) y = 2a - 1 On putting x = 2 and y = 3 in given solution. ∴ 3 2 - (a-1) 3 = 2a - 1 ⇒ 6 - 3a + 3 = 2a - 1 ⇒ - 3a ... 2 2) 3 = 3b ⇒ 10 - 9 = 3b ⇒ 1 = 3b ⇒ 1 / 3 = b Hence, the value of b is 1 / 3.

Find the value of k for 5x +2ky =3k, if x =1 and y =1 is its solution. -Maths 9th

Description : Find the value of k for 5x +2ky =3k, if x =1 and y =1 is its solution. -Maths 9th

Last Answer : Given, equation is 5x + 2ky = 3k. On putting x =1 and y =1 in this equation, we get 5(1) + 2k(1) =3k ⇒ 5 + 2k =3k ⇒ 5 = 3k - 2k ⇒ k = 5 Hence, required value of k is 5.

The value of the polynomial 5x – 4x2 + 3, when x = -1 is -Maths 9th

Description : The value of the polynomial 5x – 4x2 + 3, when x = -1 is -Maths 9th

Last Answer : (a) Let p (x) = 5x – 4x2 + 3 …(i) On putting x = -1 in Eq. (i), we get p(-1) = 5(-1) -4(-1)2 + 3= - 5 - 4 + 3 = -6

Find the value of k for 5x +2ky =3k, if x =1 and y =1 is its solution. -Maths 9th

Description : Find the value of k for 5x +2ky =3k, if x =1 and y =1 is its solution. -Maths 9th

Last Answer : Given, equation is 5x + 2ky = 3k. On putting x =1 and y =1 in this equation, we get 5(1) + 2k(1) =3k ⇒ 5 + 2k =3k ⇒ 5 = 3k - 2k ⇒ k = 5 Hence, required value of k is 5.

The value of the polynomial 5x – 4x2 + 3, when x = -1 is -Maths 9th

Description : The value of the polynomial 5x – 4x2 + 3, when x = -1 is -Maths 9th

Last Answer : (a) Let p (x) = 5x – 4x2 + 3 …(i) On putting x = -1 in Eq. (i), we get p(-1) = 5(-1) -4(-1)2 + 3= - 5 - 4 + 3 = -6

Find the value of k if (x-2)is a factor of polynomial p(x) = 2x(cube) - 6x(square) + 5x + k. -Maths 9th

Description : Find the value of k if (x-2)is a factor of polynomial p(x) = 2x(cube) - 6x(square) + 5x + k. -Maths 9th

Last Answer : Solution :-

What must be added to 2x(square) - 5x + 6 to get x(cube) - 3x(square) + 3x - 5? -Maths 9th

Description : What must be added to 2x(square) - 5x + 6 to get x(cube) - 3x(square) + 3x - 5? -Maths 9th

Last Answer : Solution :-

If x = 0 and y = k is a solution of the equation 5x - 3 y = 0, find the value of k. -Maths 9th

Description : If x = 0 and y = k is a solution of the equation 5x - 3 y = 0, find the value of k. -Maths 9th

Last Answer : Solution :-

Find the solution set of x^2 – 5x + 6 > 0. -Maths 9th

Description : Find the solution set of x^2 – 5x + 6 > 0. -Maths 9th

Last Answer : answer:

Solve x^2 – 5x + 4 > 0. -Maths 9th

Description : Solve x^2 – 5x + 4 > 0. -Maths 9th

Last Answer : answer:

What is (x^2-3x+2)/(x^2-5x +6)÷(x^2-5x+4)/(x^2-7x+12) equal to? -Maths 9th

Description : What is (x^2-3x+2)/(x^2-5x +6)÷(x^2-5x+4)/(x^2-7x+12) equal to? -Maths 9th

Last Answer : answer:

Find the value of the polynomial 5x – 4x2 + 3 at x = 2 and x = –1 -Maths 9th

Description : Find the value of the polynomial 5x – 4x2 + 3 at x = 2 and x = –1 -Maths 9th

Last Answer : Let the polynomial be f(x) = 5x – 4x2 + 3 Now, for x = 2, f(2) = 5(2) – 4(2)2 + 3 => f(2) = 10 – 16 + 3 = –3 Or, the value of the polynomial 5x – 4x2 + 3 at x = 2 ... (–1) = –5 –4 + 3 = -6 The value of the polynomial 5x – 4x2 + 3 at x = -1 is -6.

Let a function f defined from `R -> R` as `f(x)=[x+p^2 for x<=2 and px+5 , for x>2` , If the function is surjective, then find the sum of all possible

Description : Let a function f defined from `R -> R` as `f(x)=[x+p^2 for x2` , If the function is surjective, then find the sum of all possible

Last Answer : Let a function f defined from `R -> R` as `f(x)=[x+p^2 for x2` , If the function is ... of all possible integral values of p in `[-100,100]`.

Degree of the polynomial 4x4 + Ox3 + Ox5 + 5x+ 7 is -Maths 9th

Description : Degree of the polynomial 4x4 + Ox3 + Ox5 + 5x+ 7 is -Maths 9th

Last Answer : (a) Degree of 4x4 + Ox3 + Ox5 + 5x + 7 is equal to the highest power of variable x. Here, the highest power of x is 4, Hence, the degree of a polynomial is 4.

Degree of the polynomial 4x4 + Ox3 + Ox5 + 5x+ 7 is -Maths 9th

Description : Degree of the polynomial 4x4 + Ox3 + Ox5 + 5x+ 7 is -Maths 9th

Last Answer : (a) Degree of 4x4 + Ox3 + Ox5 + 5x + 7 is equal to the highest power of variable x. Here, the highest power of x is 4, Hence, the degree of a polynomial is 4.

Write a solution of the linear equation 5x + 0y +8 = 0 in two variables. -Maths 9th

Description : Write a solution of the linear equation 5x + 0y +8 = 0 in two variables. -Maths 9th

Last Answer : Solution : -

Give the geometric interpretations of 5x + 3 = 3x – 7 as an equation (i) in one variable (ii) in two variables. -Maths 9th

Description : Give the geometric interpretations of 5x + 3 = 3x – 7 as an equation (i) in one variable (ii) in two variables. -Maths 9th

Last Answer : Solution :-

For what value of k will the roots of the equation kx^2 – 5x + 6 = 0 be in the ratio 2 : 3 ? -Maths 9th

Description : For what value of k will the roots of the equation kx^2 – 5x + 6 = 0 be in the ratio 2 : 3 ? -Maths 9th

Last Answer : (b) 1 Let the roots of the equation kx2 - 5x + 6 = 0 be α and β. Then, α + β = \(\frac{5}{k}\) ...(i) αβ = \(\frac{6}{k}\) ...(ii) Given \(\ ... frac{9}{k}\) ⇒ 9k2 - 9k = 0 k(k - 1) = 0 ⇒ k = 0 or 1 But k = 0 does not satisfy the condition, so k = 1.

What is the equation of the line having the y-intercept –1 and parallel to the line y = 5x – 7 ? -Maths 9th

Description : What is the equation of the line having the y-intercept –1 and parallel to the line y = 5x – 7 ? -Maths 9th

Last Answer : Slope of AB = \(rac{2-4}{1-0}\) = -2, Slope of BC = \(rac{3-2}{3-1}\) = \(rac{1}{2}\)Slope of AC = \(rac{3-4}{3-0}\) = \(-rac{1}{3}\)Slope of AB × Slope of BC = -2 x \(rac{1}{2}\) = -1∴ AB ⊥ BC, i.e, ∠B = 90º ⇒ ΔABC is a right angled.

Find minimum value of `px+qy` where `p>0, q>0, x>0, y>0` when `xy=r,^2` without using derivatives.

Description : Find minimum value of `px+qy` where `p>0, q>0, x>0, y>0` when `xy=r,^2` without using derivatives.

Last Answer : Find minimum value of `px+qy` where `p>0, q>0, x>0, y>0` when `xy=r,^2` without using ... `pqsqrtr` B. `2pqsqrtr` C. `2rsqrtpq` D. None of these

If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Description : If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Last Answer : (d) We know that, a point lies on X-axis, if its y-coordinate is zero. So, on plotting the given points on graph paper, we get Q and O lie on the X-axis.

If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Description : If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Last Answer : (d) We know that, a point lies on X-axis, if its y-coordinate is zero. So, on plotting the given points on graph paper, we get Q and O lie on the X-axis.

Let R be a relation on the set N, defined by {(x, y) : 2x – y = 10} then R is -Maths 9th

Description : Let R be a relation on the set N, defined by {(x, y) : 2x – y = 10} then R is -Maths 9th

Last Answer : (a) ReflexiveGiven, {(\(x\), y) : 2\(x\) – y = 10} Reflexive, \(x\) R \(x\) = 2\(x\) – \(x\) = 10 ⇒ \(x\) = 10 ⇒ y = 10 ∴ Point (10, 10) ∈ N ⇒ R is reflexive.

Find the slope of the line whose equation is px - qy = r + 1.?

Description : Find the slope of the line whose equation is px - qy = r + 1.?

Last Answer : 7

If both (x+1) and (x -1) are factors of ax3 + x2 - 2x + b , find a and b. -Maths 9th

Description : If both (x+1) and (x -1) are factors of ax3 + x2 - 2x + b , find a and b. -Maths 9th

Last Answer : Let p(x) = ax3 + x2 - 2x + b Since (x+1) and (x-1) are the factors of p(x), ∴ p(-1) = 0 and p(1) = 0 ∴ p(-1) = a(-1)3 + (-1)2 - 2 (-1) + b = 0 ⇒ - a + 1 + 2 + b = 0 ⇒ a - b = 3 ---- (i) ... 0 ⇒ a + 1 - 2 + b = 0 ⇒ a + b = 1 ----- (ii) solving equations (i) and (ii) we get a = 2 and b = -1

If both (x+1) and (x -1) are factors of ax3 + x2 - 2x + b , find a and b. -Maths 9th

Description : If both (x+1) and (x -1) are factors of ax3 + x2 - 2x + b , find a and b. -Maths 9th

Last Answer : Let p(x) = ax3 + x2 - 2x + b Since (x+1) and (x-1) are the factors of p(x), ∴ p(-1) = 0 and p(1) = 0 ∴ p(-1) = a(-1)3 + (-1)2 - 2 (-1) + b = 0 ⇒ - a + 1 + 2 + b = 0 ⇒ a - b = 3 ---- (i) ... 0 ⇒ a + 1 - 2 + b = 0 ⇒ a + b = 1 ----- (ii) solving equations (i) and (ii) we get a = 2 and b = -1

If a, b are the roots of the equation `x^(2) - px +q = 0`, then find the equation which has `a/b` and `b/a` as its roots.

Description : If a, b are the roots of the equation `x^(2) - px +q = 0`, then find the equation which has `a/b` and `b/a` as its roots.

Last Answer : If a, b are the roots of the equation `x^(2) - px +q = 0`, then find the equation which has `a/b` and `b/a` as its roots.

If α , β are the zeroes of f(x) = px 2 – 2x + 3p and α + β = αβ then the value of p is: (a) 1/3 (b) -2/3 (c) 2/3 (d) -1/3

Description : If α , β are the zeroes of f(x) = px 2 – 2x + 3p and α + β = αβ then the value of p is: (a) 1/3 (b) -2/3 (c) 2/3 (d) -1/3

Last Answer : (c) 2/3

If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Description : If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Last Answer : Join PR. ∵ △PSR and ||gm APRD are on the same base and between same parallel lines. ar(△PSR) = 1/2 ar(||gm APRD) Similarly, ar(△PQR) = 1/2 ar(||gm PBCR) ar(△PQRS) = ar(△PSR) + △(PQR) = 1/2 ar(||gm APRD) + 1 ... |gm PBCR) = 1/2 ar(||gm ABCD) ⇒ ar(||gm ABCD) = 2 ar(||gm PQRS) = 2 32.5 = 65cm2

The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Description : The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Last Answer : Let r be the radius, then OQ = OB = r and OR = (r - 4) ∴ OQ2 = OR2 + RO2 ⇒ r2 = 64 + (r-4)2 ⇒ r2 = 64 + r2 + 16 - 8r ⇒ 8r = 80 ⇒ r = 10 cm

If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Description : If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Last Answer : (b) In point P (-1, 1), x-coordinate is -1 unit and y-coordinate is 1 unit, so it lies in llnd quadrant. Similarly, we can plot all the points Q (3, -4), R (1, -1), S (-2, -3) and T (-4, 4), It is clear from the graph that points R and Q lie in fourth quadrant.