Answer:
[tex]y = \times {}^{2} + 2 \times {}^{2} + \times + 5[/tex]
To find x - intercept/zero, substitute y = 0
[tex]0 = \times {}^{2} + 2 \times {}^{2} + \times + 5[/tex]
Swap the sides of the equation
[tex] \times {}^{2} + 2 \times {}^{2} + \times + 5 = 0 [/tex]
Collect the terms
[tex]3 \times {}^{2} + \times + 5 = 0[/tex]
Solve the quadratic equation
[tex]ax {}^{2} + bx \: + c = 0 \: using[/tex]
[tex]x = - b + \sqrt{b {}^{2} - 4ac} \\ 2a[/tex]
[tex]x = - 1 + \sqrt{1 {}^{2} - 4 \times 3 \times 5} \: \\ 2 \times 3[/tex]
1 raised to any power equals 1
[tex]x = - 1 + \sqrt{1 - 4 \times 3 \times 5} \\ 2 \times 3[/tex]
Calculate the product
[tex]x = - 1 + \sqrt{1 - 60} \\ 2 \times 3[/tex]
Multiply the numbers
[tex]x = - 1 + \sqrt{1 - 60} \\ 6[/tex]
Calculate the difference
[tex]x = - 1 + \sqrt{ - 59} \\ 6[/tex]
The square root of a negative number does not exist in the set of real numbers.
Since there is no solution for y = 0, there is no x-intercept/Zero.
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