To solve the given problem, we need to find the value of \( k \) for which the sum of the squares of the roots of the polynomial \( x^2 + 8x + 15k = 0 \) equals 34.
Let the roots of the polynomial be \( \alpha \) and \( \beta \).
According to Vieta's formulas for a quadratic equation \( ax^2 + bx + c = 0 \):
- The sum of the roots, \( \alpha + \beta \), is given by \( -\frac{b}{a} \). Here, \( b = 8 \) and \( a = 1 \), so:
The product of the roots, \( \alpha \beta \), is given by \( \frac{c}{a} \). Here, \( c = 15k \) and \( a = 1 \), so:
\[\alpha \beta = \frac{15k}{1} = 15k\]The sum of the squares of the roots is given by:
\[\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\]Substituting the known values:
\[\alpha^2 + \beta^2 = (-8)^2 - 2(15k)\]Given that the sum of the squares of the roots is 34, we equate and solve for \( k \):
\[64 - 30k = 34\]- Subtract 64 from both sides:
Simplify to find \( k \):
\[k = \frac{-30}{-30} = 1\]The correct value of \( k \) is 1. Therefore, the correct answer is
1
.
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Important Questions from Polynomials
If (x + y) 3+ 8 (x - y) 3= (3x + Ay) (3x 2+ Bxy + Cy 2), then the value of A + B + C is:
Given that x 8- 34x 4+ 1 = 0, x > 0. What is the value of (x 3+ x -3 )?
If \(x - \frac 3 x = 6,\; x \ne 0,\) then the value of \(\frac {x^4 - \frac {27}{x^2}}{x^2 - 3x - 3}\) is:
If \(x\left(3 - \frac 2 x\right) = \frac 3 x,\) then the value of \(x^3 - \frac 1 {x^3}\) is equal to:
The coefficient of x in (x – 3y) 3is:
