Answer:
1. Two solutions
2. One solution
Step-by-step explanation:
1. 7^2-5*(-4)*8 > 0 (is a number +)
2. Is (x-5)^2=0 ---> x=5
The equation 0=-4x^2+7x+8 has two solutions and the equation 0=2x^2-20x+50 has one solution.
A quadratic equation is a second-degree algebraic equation represented by the:
[tex]\rm ax^2+bx+c=0[/tex] ........(1)
Where a, b, and c are the constants and [tex]\rm a \neq 0[/tex]
We have a quadratic equation:
[tex]\rm 0 = -4x^2+7x+8[/tex] or
[tex]\rm -4x^2+7x+8=0[/tex]
Compare this equation with the standard form of quadratic equation:
a = -4, b = 7, and c = 8
For the nature of the solution(roots):
[tex]\rm D = b^2-4ac[/tex]
[tex]\rm D = 7^2-4(-4)(8)[/tex]
D = 49 +128 ⇒177
Here D>0 which means the solutions are real and the equation has two distinct solutions.
For the quadratic equation [tex]\rm 0=2x^2-20x+50[/tex] or
[tex]\rm 2x^2-20x+50=0[/tex]
Again compare the above equation with standard equation, we get:
a = 2, b = -20, and c = 50
[tex]\rm D = (-20)^2-4(2)(50)[/tex]
D = 400 - 400 ⇒ 0
Here D = 0 which means solutions are real but equal.
Thus, the equation 0=-4x^2+7x+8 has two solutions and the equation 0=2x^2-20x+50 has one solution.
Learn more about quadratic equations here:
brainly.com/question/2263981
