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2 years ago

PLEASE HELP ASAP!!!! SUPER URGENT

Mathematics

1 answer:

DochEvi [55]2 years ago

6 0

Answer:

y = 2(x + 3)(x + 1)

Step-by-step explanation:

Use the formula y = a(x-b)(x-c)

b and c are the x-intercepts

The x-intercepts in this function are -3 and -1

y = a(x+3)(x+1)

substitute the point (-2, -1)

-2 = a(-2 + 3)(-2 + 1)

-2 = a(1)(-1)

-2 = -a

a = 2

y = 2(x + 3)(x + 1)

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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:

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$b^2 - 4ac = 2^2 - 4(1)(5) = 4 - 20 = -16$

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,

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

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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