Answer:
2 real solutions
Step-by-step explanation:
Remember this messy thing?
The <em>quadratic formula</em>, as it's called, gives us the roots to any quadratic equation in standard form (ax² + bx + c = 0). The information on the <em>type</em> of roots is contained entirely in that bit under the square root symbol (b² - 4ac), called the <em>discriminant</em>. If it's non-negative, we'll have <em>real</em> roots, if it's negative, we'll have <em>complex roots</em>.
For our equation, we have a discrimant of (-3)² - 4(6)(-4) = 9 + 96 = 105, which is non-negative, so we'll have real solutions, and since quadratics are degree 2, we'll have exactly 2 real solutions.
No solution of the system of equations y = -2x + 5 and -5y = 10x + 20 ⇒ 2nd answer
Step-by-step explanation:
Let us revise the types of solutions of a system of linear equations
- One solution
- No solution when the coefficients of x and y in the two equations are equal and the numerical terms are different
- Infinitely many solutions when the coefficients of x , y and the numerical terms are equal in the two equations
∵ y = -2x + 5
- Add 2x to both sides
∴ 2x + y = 5 ⇒ (1)
∵ -5y = 10x + 20
- Subtract 10x from both sides
∴ -10x - 5y = 20
- Divide both sides by -5
∴ 2x + y = -4 ⇒ (2)
∵ The coefficient of x in equation (1) is 2
∵ The coefficient of x in equation (2) is 2
∴ The coefficients of x in the two equations are equal
∵ The coefficient of y in equation (1) is 1
∵ The coefficient of y in equation (2) is 1
∴ The coefficients of y in the two equations are equal
∵ The numerical term in equation (1) is 5
∵ The numerical term in equation (2) is -4
∴ The numerical terms are different
From the 2nd rule above
∴ No solution of the system of equations
No solution of the system of equations y = -2x + 5 and -5y = 10x + 20
Learn more:
You can learn more about the system of equations in brainly.com/question/6075514
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Answer:
its an odd
Step-by-step explanation:
The leading coefficient is negative
hope this helps
427.26 yards- I am pretty sure.
