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Which parabola will have one real solution with the line y = x – 5? y = x2 x – 4 y = x2 2x – 1 y = x2 6x 9 y = x2 7x 4

Sagot :

System of equations y = x² + 7x + 4 has one real root with the line y = x -5.

What is the discriminant in the quadratic equation?

A quadratic equation is the polynomial equation of two degrees in one variable. The general form of a quadratic equation is  ax² + bx + c = 0.

The quadratic equation formula for the solution of roots.

[tex]\rm (\alpha ,\beta ) = [ -b[/tex]±[tex]\rm \sqrt{b^{2} - 4ac} /2ac[/tex]]

The discriminant is the [tex]\rm \sqrt{b^{2} - 4ac}[/tex] part of the quadratic equation

If D is greater than zero then the roots are real and distinct.

If D is equal to zero then the roots are real and equal.

If D is less than zero then the roots are imaginary and unequal.

y = x² + 7x + 4

From First option, System of equations

y = x² + x – 4

y = x – 5

So, x - 5 = x² + x – 4

x² + x – 4 - x + 5 = 0

x² + 1 = 0

discriminant = b²- 4×a×c

= 0²- 4×1×1

-4 < 0, therefore the equation has no real roots.

From Second option, System of equations

y = x² + 2x – 1

y = x – 5

So, x - 5 = x² + 2x – 1

x² + 2x - 1  - x + 5 = 0

x² + x + 4 = 0

discriminant = b²- 4×a×c

= 1²-4×1×4

-15 < 0   therefore, the equation has no real roots.

From Third option, System of equations

y = x² + 6x + 9

y = x – 5

So, x - 5 = x² + 6x + 9

x² + 6x + 9  - x + 5 = 0

x² + 5x + 4 = 0

discriminant = b²- 4×a×c

= 5²-4×1×4

9 > 0 therefore, the equation has two different real roots.

From Fourth option, System of equations

y = x² + 7x + 4

y = x – 5

So. x - 5 = x² + 7x + 4

x² + 7x + 4 - x + 5 = 0

x² + 6x + 9 = 0

discriminant = b²- 4×a×c

= 6²-4×1×9 = 0

0  therefore, the equation has one real root.

Since, System of equations y = x² + 7x + 4 has one real root with the line y = x -5.

Learn more about quadratic equations;

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Answer:

D on edge

Step-by-step explanation: