Solve the system of equations. y = 2x^2 - 42x + 4 y = -42x + 36 Write the coordinates in exact form. Simplify all fractions and radicals. (______,______) (______,______)

Set Equations Equal: Set the two equations equal to each other since they both equal y. y = 2x^2 - 42x + 4 y = -42x + 36 2x^2 - 42x + 4 = -42x + 36Simplify Quadratic Equation: Simplify the equation by moving all terms to one side to form a quadratic equation. 2x^2 - 42x + 4 + 42x - 36 = 0 2x^2 - 32 = 0Solve for x: Solve the quadratic equation for x. 2x^2 = 32 x^2 = 16 x = ±4Substitute for y: Find the corresponding y-values for each x-value by substituting back into one of the original equations. We'll use y = -42x + 36. For x = 4: y = -42(4) + 36 y = -168 + 36 y = -132Find Second y-Value: Find the corresponding y-value for the second x-value. For x = -4: y = -42(-4) + 36 y = 168 + 36 y = 204Write Coordinates: Write the coordinates in exact form. First Coordinate: (4, -132) Second Coordinate: (-4, 204)

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Solve the system of equations. $\newline$ $y = 2x^2 - 42x + 4$ $\newline$ $y = -42x + 36$ $\newline$ Write the coordinates in exact form. Simplify all fractions and radicals. $\newline$ (,) $\newline$ (,)

View step-by-step help

Home

Math Problems

Algebra 2

Solve a system of linear and quadratic equations: parabolas

Full solution

Q. Solve the system of equations.

\newline

y = 2x^2 - 42x + 4

\newline

y = -42x + 36

\newline

Write the coordinates in exact form. Simplify all fractions and radicals.

\newline

(______,______)

\newline

(______,______)

Set Equations Equal: Set the two equations equal to each other since they both equal $y$ . $y = 2x^2 - 42x + 4$ $y = -42x + 36$ $2x^2 - 42x + 4 = -42x + 36$
Simplify Quadratic Equation: Simplify the equation by moving all terms to one side to form a quadratic equation. $\newline$ $2x^2 - 42x + 4 + 42x - 36 = 0$ $\newline$ $2x^2 - 32 = 0$
Solve for x: Solve the quadratic equation for x. $\newline$ $2x^2 = 32$ $\newline$ $x^2 = 16$ $\newline$ $x = \pm4$
Substitute for y: Find the corresponding $y$ -values for each $x$ -value by substituting back into one of the original equations. We'll use $y = -42x + 36$ . $\newline$ For $x = 4$ : $\newline$ $y = -42(4) + 36$ $\newline$ $y = -168 + 36$ $\newline$ $y = -132$
Find Second y-Value: Find the corresponding y-value for the second x-value. $\newline$ For $x = -4$ : $\newline$ $y = -42(-4) + 36$ $\newline$ $y = 168 + 36$ $\newline$ $y = 204$
Write Coordinates: Write the coordinates in exact form. $\newline$ First Coordinate: $(4, -132)$ $\newline$ Second Coordinate: $(-4, 204)$