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what is the y-intercepr of the equation y=×^2+2×^2 + ×+5​

Sagot :

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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