Question
4. How many real-number solutions does the equation have?
`x^{2}+7x-5=0`
`x^{2}+7x-5=0`
Answers
Answered by
GPT 3.5
To determine the number of real-number solutions for the equation `x^{2}+7x-5=0`, we need to calculate the discriminant of the quadratic equation which is given by `b^{2}-4ac`.
In this case, `a=1, b=7,` and `c=-5`.
Therefore, the discriminant (`D`) can be calculated as follows:
`D = b^{2}-4ac = (7)^{2}-4(1)(-5) = 49+20 = 69`.
Since the discriminant (`D`) is positive and greater than zero (`D>0`), the quadratic equation `x^{2}+7x-5=0` has two distinct real solutions.
Therefore, the equation `x^{2}+7x-5=0` has 2 real-number solutions.
In this case, `a=1, b=7,` and `c=-5`.
Therefore, the discriminant (`D`) can be calculated as follows:
`D = b^{2}-4ac = (7)^{2}-4(1)(-5) = 49+20 = 69`.
Since the discriminant (`D`) is positive and greater than zero (`D>0`), the quadratic equation `x^{2}+7x-5=0` has two distinct real solutions.
Therefore, the equation `x^{2}+7x-5=0` has 2 real-number solutions.
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