Respuesta :
Question:
Find the number of real number solutions for the equation. x^2 + 5x + 7 = 0
Answer:
The number of real solutions for the equation [tex]x^{2}+5 x+7=0[/tex] is zero
Solution:
For a Quadratic Equation of form : [tex]a x^{2}+b x+c=0[/tex] ---- eqn 1
The solution is [tex]x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}[/tex]
Now , the given Quadratic Equation is [tex]x^{2}+5 x+7=0[/tex] ---- eqn 2
On comparing Equation (1) and Equation(2), we get
a = 1 , b = 5 and c = 7
In [tex]x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}[/tex] , [tex]b^2 - 4ac[/tex] is called the discriminant of the quadratic equation
Its value determines the nature of roots
Now, here are the rules with discriminants:
1) D > 0; there are 2 real solutions in the equation
2) D = 0; there is 1 real solution in the equation
3) D < 0; there are no real solutions in the equation
Now let solve for given equation
[tex]D= b^2 - 4ac\\\\D = 5^2 - 4(1)(7)\\\\D = 25 - 28 \\\\D = -3[/tex]
Since -3 is less than 0, this means that there are 0 real solutions in this equation.
