Respuesta :
A) No solutions, because the number under the square root symbol in the quadratic equation (b^2 - 4ac) is negative.
B) One solutions, because the number under the square root symbol in the quadratic equation (b^2 - 4ac) is equal to 0.
B) One solutions, because the number under the square root symbol in the quadratic equation (b^2 - 4ac) is equal to 0.
Answer: The correct options are
(1) (a) No solutions.
(2) (b) One solution.
Step-by-step explanation: We are given to find the number of real solutions to the following two quadratic equations:
[tex]-4x^2+7x-8=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\3x^2+18x+27=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)[/tex]
We know that
for quadratic equation [tex]ax^2+bx+c=0~a\neq 0,[/tex] the type of solution depends on the discriminant [tex]D=b^2-4ac[/tex] as follows:
(i) there are two real solutions if D is greater than 0.
(ii) there is one real solution if D is equal to 0.
(ii) there is no real solution if D is less than 0.
For equation (i), we have
a = -4, b = 7 and c = 8.
Therefore, the discriminant is given by
[tex]D=b^2-4ac=7^2-4\times (-4)(-8)=79-148=-69<0.[/tex]
So, there will be no real solution.
For equation (ii), we have
a = 3, b = 18 and c = 27.
Therefore, the discriminant is given by
[tex]D=b^2-4ac=18^2-4\times 3\times 27=324-324=0.[/tex]
So, there will be one real solution.
Thus, the correct options are
(1) (a) No solutions.
(2) (b) One solution.