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Chapter 3: Problem 177

If one root of the equation \(a X^{2}-6 x+c+9=0\) \((a, c \in R, a \neq 0)\) is \(3-5 i\) then \(a=\) (a) 1,25 (b) \(-1,25\) (c) \(1,-25\) (d) \(-1,-25\)

Short Answer

Expert verified
The correct value of \(a\) is 1, and the answer is option (a), i.e., \(a = 1,25\).

Step by step solution

01

Find the conjugate of the given root

Since one root of the equation is \(3-5i\), the conjugate of this root will also be a root of the equation. The conjugate of a complex number \(a+bi\) is \(a-bi\). Therefore, the conjugate of \(3-5i\) is \(3+5i\).
02

Use the sum and product of roots

For any quadratic equation of the form \(Ax^2 + Bx + C = 0\), the sum of its roots is \(-\frac{B}{A}\) and the product of its roots is \(\frac{C}{A}\). In our case, it can be written as: Sum of roots: \(3-5i + 3+5i = -\frac{-6}{a}\) and Product of roots: \((3-5i)(3+5i) = \frac{c+9}{a}\).
03

Calculate the sum of roots

Compute the sum of the roots: \(3-5i + 3+5i=-\frac{-6}{a}\), \(-\frac{-6}{a} = 6\). Solve for \(a\): \(a = 1\).
04

Calculate the product of roots

The product of the roots: \((3-5i)(3+5i)=\frac{c+9}{1}\) (since we found the value of \(a\) to be 1). Use the property for the product of conjugates: \((3-5i)(3+5i)=3^2 - (5i)^2=9+25=34\). So, \(c+9 = 34\).
05

Find the value of a

We have already found the value of \(a\) in Step 3. The correct value of \(a\) is \(a = 1\). Checking the given options, we find that the answer is option (a), i.e., \(a = 1,25\).

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Most popular questions from this chapter

If the roots of the quadratic equation \((2 \mathrm{k}+3) \mathrm{x}^{2}+2(\mathrm{k}+3)\) \(x+(k+5)=0\\{k \in R, k \neq\\{(-3) / 2\\}\\}\) are equal, then \(\mathrm{K}=\) (a) 1,6 (b) \(-1,-6\) (c) \(-1,6\) (d) \(1,-6\)

The minimum value of \((\mathrm{x}+\mathrm{a})^{2}+(\mathrm{x}+\mathrm{b})^{2}+(\mathrm{x}+\mathrm{c})^{2}\) will be at \(\mathrm{x}\) equal to (a) \([(a+b+c) / 3]\) (b) \([\\{-(a+b+c)\\} / 3]\) (c) \(\sqrt{(a b c)}\) (d) \(a^{2}+b^{2}+c^{2}\)

The number of values of \(\mathrm{x}\) in the interval \([0,3 \pi]\) Satisfying the equation \(2 \sin ^{2} \mathrm{x}+5 \sin \mathrm{x}-3=0\) is (a) 6 (b) 1 (c) 2 (d) 4

If the equation \(x^{2}-m(2 x-8)-15=0\) has equal roots then \(\mathrm{m}=\) (a) \(3,-5\) (b) \(-3,5\) (c) 3,5 (d) \(-3,-5\)

If \(\alpha \& \beta\) are the roots of the equation \(x^{2}-x+1=0\) then \(\alpha^{2009}+\beta^{2009}=\) (a) \(-1\) (b) 1 (c) \(-2\) (d) 2
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