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Chapter 3: Problem 194

For equation \(2 \mathrm{x}^{2}+16 \mathrm{x}+3 \mathrm{k}=0\) sum of the squares of roots is 10 then \(\mathrm{k}=\) (a) 12 (b) 15 (c) 18 (d) 21

Short Answer

Expert verified
The value of k is 18, which corresponds with answer choice (c).

Step by step solution

01

Find sum and product of roots

Following Vieta's formulas, for a quadratic equation of the form \(ax^2 + bx +c = 0\), the sum of its roots is given by \(-\frac{b}{a}\) and the product of its roots is given by \(\frac{c}{a}\). Here, our equation is \(2x^2 + 16x + 3k = 0\), where \(a = 2\), \(b = 16\), and \(c = 3k\). So, the sum of roots is: \(-\frac{16}{2} = -8\) And, the product of roots is: \(\frac{3k}{2}\) Let the roots be α and β. Thus, \(\alpha + \beta = -8\) and \(\alpha\beta = \frac{3k}{2}\).
02

Write sum of squares of roots in terms of α and β

We are given that the sum of squares of roots is 10. In terms of α and β, this can be expressed as: \(\alpha^2 + \beta^2 = 10\) We can rewrite the above expression in terms of the sum and product of roots: \((\alpha + \beta)^2 - 2\alpha\beta = 10\)
03

Substitute the values for sum and product of roots

Substitute the values of \(\alpha + \beta = -8\) and \(\alpha\beta = \frac{3k}{2}\) in the above equation: \((-8)^2 - 2\left(\frac{3k}{2}\right) = 10\) Now, we have an equation with k as the only unknown variable.
04

Solve the equation for k

Simplify the equation and solve for k: \(64 - 3k = 10\) Add 3k and subtract 10 from both sides: \(64 - 10 = 3k\) \(54 = 3k\) Divide by 3: \(k = 18\) The value of k is 18, which corresponds with answer choice (c).

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Most popular questions from this chapter

If \(0 \leq \mathrm{x} \leq \pi\) and \(16^{(\sin ) 2 \mathrm{x}}+16^{(\cos ) 2 \mathrm{x}}=10\) then \(\mathrm{x}=\) (a) \((\pi / 3)\) (b) \((\pi / 2)\) (c) \((\pi / 4)\) (d) \((3 \pi / 4)\)

If \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are rational numbers then the roots of equation \(a b c^{2} x^{2}+3 a^{2} c x+b^{2} c x-6 a^{2}-a b+2 b^{2}=0\) are \(\ldots \ldots \ldots\) (a) imaginary (b) equals (c) rational (d) irrational

If roots of equation \(x^{2}-2 a x+a^{2}+a-3=0\) are real and less than 3 then........... (a) \(a<2\) (b) \(2 \leq \mathrm{a} \leq 3\) (c) \(3<\mathrm{a} \leq 4\) (d) \(a>4\)

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