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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 \(\alpha, \beta\) are roots of equation \(\mathrm{x}^{2}-\mathrm{p} \mathrm{x}+\mathrm{r}=0\) and \((\alpha / 2), \alpha \beta\) are roots of equation \(\mathrm{x}^{2}-\mathrm{qx}+\mathrm{r}=0\) then \(\mathrm{r}=\ldots \ldots\) (a) \((2 / 9)(\mathrm{p}-\mathrm{q})(2 \mathrm{q}-\mathrm{p})\) (b) \((2 / 9)(q-p)(2 p-q)\) (c) \((2 / 9)(q-2 p)(2 q-p)\) (d) \((2 / 9)(2 \mathrm{p}-\mathrm{q})(2 \mathrm{q}-\mathrm{p})\)

For the equation \(\mathrm{x}^{2}+\mathrm{k}^{2}=(2 \mathrm{k}+2) \mathrm{x}, \mathrm{k} \in \mathrm{R}\) roots are complex then (a) \(\overline{k=\\{(-1) / 2\\}}\) (b) \(k>\\{(-1) / 2\\}\) (c) \(k>\\{(-1) / 2\\}\) (d) \(\\{(-1) / 2\\}<\mathrm{k}<0\)

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 \(\mathrm{f}(\mathrm{x})=\mathrm{x}-[\mathrm{x}], \mathrm{x} \in \mathrm{R}-\\{0\\}\) where \([\mathrm{x}]=\) the greatest integer not greater than \(\mathrm{x}\), than number of solution \(\mathrm{f}(\mathrm{x})+\mathrm{f}(1 / \mathrm{x})=1 \ldots \ldots \ldots \ldots \ldots\) (a) 0 (b) 1 (c) 2 (d) infinite

If the product of roots of equation \(x^{2}-5 k x+4 e^{4 \log k}-3=0\) is 61 then \(\mathrm{k}\) is \(\ldots \ldots \ldots \ldots\) (a) 1 (b) 2 (c) 3 (d) 4
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