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If one zero is the reciprocal of other find the value of k in 3kx² - 7x + 11

Posted by Shristy Dewangan 2 years, 3 months ago

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The product of the zeros of a quadratic equation is equal to the constant term divided by the coefficient of the squared term. So, for the quadratic equation 3kx² - 7x + 11, the product of the zeros would be: x * (1/x) = 1 Now, we can use the sum of the zeros of a quadratic equation, which is equal to the negative coefficient of the linear term divided by the coefficient of the squared term. So, for our equation: x + (1/x) = -(-7) / 3k x + (1/x) = 7 / 3k Now, we have two equations: x * (1/x) = 1 x + (1/x) = 7 / 3k Let's simplify the first equation: x * (1/x) = 1 1 = 1 Now, for the second equation, let's solve for x + (1/x): x + (1/x) = 7 / 3k Since we know that x * (1/x) = 1, we can multiply both sides of the second equation by x: x * (x + (1/x)) = x * (7 / 3k) This simplifies to: x² + 1 = 7x / 3k Now, let's find the common denominator on the right side: x² + 1 = (7x * x) / (3k * x) x² + 1 = (7x²) / (3kx) Now, let's bring all terms to one side of the equation: x² + 1 - (7x²) / (3kx) = 0 To solve this quadratic equation for x, we need to find the common denominator: (3kx * x² + 3kx - 7x²) / (3kx) = 0 Now, we need to find the LCD (Least Common Denominator) of the terms in the numerator: LCD = 3kx Now, let's multiply both sides of the equation by the LCD to get rid of the fraction: 3kx * (3kx * x² + 3kx - 7x²) / (3kx) = 0 * (3kx) This simplifies to: 3kx * (3kx * x² + 3kx - 7x²) = 0 Expanding the expression: 9k²x³ + 3k²x² - 21kx³ = 0 Combine the x³ terms: (9k² - 21k) * x³ + 3k²x² = 0 For this equation to be true for all x, the coefficients of x³ and x² must be zero: 9k² - 21k = 0 (Coefficient of x³) 3k² = 0 (Coefficient of x²) Solving the second equation: 3k² = 0 k² = 0 k = 0 Now, for the first equation: 9k² - 21k = 0 9(0)² - 21(0) = 0 0 - 0 = 0 Since both equations are true when k = 0, the value of k that satisfies the conditions is k = 0.

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A teacher had written on the blackboard the following algebraic expressions: 1. 2x + 15 2. x^2 + 6x 3. x^2 + 8x - 15 4. x^3 - 12x + 7 The teacher called two students, Sita and Kamal. I. The teacher told Sita to add expressions (1) and (2). II. Now the teacher said Kamal to subtract expression (2) from (3). Now answer the following questions: i. Find the zeros of the polynomial formed by Kamal. ii. If a and b are zeros of the polynomial formed by Sita, then find the value of ab.

Posted by Traptijain Jain 2 years, 3 months ago

CBSE > Class 10 > Mathematics

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What is mean by identity

Posted by Shreyan Anarase 2 years, 3 months ago

CBSE > Class 10 > Mathematics

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1 ÷ 2 alpha + beta and 1 ÷ 2alpha + beta

Posted by Shwetha P 2 years, 3 months ago

CBSE > Class 10 > Mathematics

2 answers

Harsh Vardhan 2 years, 3 months ago

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Adwaith K 2 years, 3 months ago

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Card total table

Posted by Shivendra Pal 2 years, 3 months ago

CBSE > Class 10 > Mathematics

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5^(x+1)+5^(2-x)=5^3+1 by factorisation

Posted by Ashutosh Das 2 years, 3 months ago

CBSE > Class 10 > Mathematics

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The line segment joining A(-2,9) and B(6,3) is a diameter of a circle with centre C. Fin the coordinates of C.

Posted by Ragini Shaw 2 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Haret Gujjar 2 years, 3 months ago

C(2,6)

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In DE || BC then the value of EC is

Posted by Jay Sutar 2 years, 3 months ago

CBSE > Class 10 > Mathematics

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Haret Gujjar 2 years, 3 months ago

Ec equal BD

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Find all the zeros of polynomial f(x) = 2x⁴- 3x³ - 3x² + 6x -2 , if being given that two of its zero are √2 and -√2

Posted by Yogita Modi 2 years, 3 months ago

CBSE > Class 10 > Mathematics

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if y=e^acos-1x , -1<x<1, show that (1-x^2) d^2y/dx^2-xdy/dx-a^2y=0

Posted by Manvi Manvi 2 years, 3 months ago

CBSE > Class 10 > Mathematics

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If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289,find sum of first n terms

Posted by Rishi Kumar Sandaka 2 years, 3 months ago

CBSE > Class 10 > Mathematics

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Xyz Xyz 2 years, 3 months ago

7=49=7^2 17=289=17^2 n=n^2

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9x-10y=-12 equation first 2x+3y=13 second equation solve it by substitution method

Posted by Akshita Nahar 2 years, 3 months ago

CBSE > Class 10 > Mathematics

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Shrishti Mishra 2 years, 3 months ago

9x-10y=-12 -eq.1 2x+3y=13 -eq.2 Lets take eq2 2x+3y=13 2x=13-3y X=13-3y/2 eq 3 Now let's place the value of x from eq 3 to eq 1 9x-10y=-12 9(13-3y/2)-10y=-12 117-27y-20y=-24 47y=-24-117 47y=141 Y=141/47 Y=3 -eq4 Putting the value of y through eq4 in eq2 2x+3y=13 2x+3(3)=13 2x+9=13 2x=13-9 2x=4 X=4/2 X=2 - eq5 Hence from equation 4 and 5 we can say that the value of x=2 and the value of y=3 - Hence proved Hope it helps

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3x+4y=10

Posted by Khwaish Gangwani 2 years, 3 months ago

CBSE > Class 10 > Mathematics