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4 years ago

One to one functions g and h are defined as follows.

1 answer:

6 0

1. g^-1 ( - 1 ) = 6

2. h^-1 ( x ) = ?
y = 2 x - 9
- 2 x = - y - 9 / : ( -2 )
x = ( y + 9 )/ 2
h^-1 ( x ) = ( x + 9 ) / 2

3. ( h ° h^-1) ( 1 ) = ?
h ( h^-1 ( 1 ) )= h ( (1+9) /2 ) = h ( 5 ) = 2· 5 - 9 = 10 - 9 = 1

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nirvana33 [79]

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3 years ago

What are the solutions of x^2-2x+5=0

sweet [91]

The solution of $x^{2}-2 x+5=0$ are 1 + 2i and 1 – 2i

Solution:

Given, equation is $x^{2}-2 x+5=0$

We have to find the roots of the given quadratic equation

Now, let us use the quadratic formula

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$ --- (1)

Let us determine the nature of roots:

Here in $x^{2}-2 x+5=0$ a = 1 ; b = -2 ; c = 5

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Since $b^2 - 4ac < 0$ , the roots obtained will be complex conjugates.

Now plug in values in eqn 1, we get,

$x=\frac{-(-2) \pm \sqrt{(-2)^{2}-4 \times 1 \times 5}}{2 \times 1}$

On solving we get,

$x=\frac{2 \pm \sqrt{4-20}}{2}$

$x=\frac{2 \pm \sqrt{-16}}{2}$

$x=\frac{2 \pm \sqrt{16} \times \sqrt{-1}}{2}$

we know that square root of -1 is "i" which is a complex number

$\begin{array}{l}{\mathrm{x}=\frac{2 \pm 4 i}{2}} \\\\ {\mathrm{x}=1 \pm 2 i}\end{array}$

Hence, the roots of the given quadratic equation are 1 + 2i and 1 – 2i

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3 years ago

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mezya [45]

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3 years ago

Read 2 more answers