Question

Use the Quadratic Formula to find the exact solutions of x2 + 7x − 4 = 0. x equals negative b plus or minus the square root of b squared minus 4 times a times c, all over 2 times a x equals negative 7 plus or minus the square root of 33, all over 2 x equals negative 7 plus or minus the square root of 65, all over 2 x equals 7 plus or minus the square root of 65, all over 2 x equals 7 plus or minus the square root of 33, all over 2

Answer 1

Answer:

Option 2

x equals negative 7 plus or minus the square root of 65, all over 2

Step-by-step explanation:

x² + 7x - 4 = 0

x = [-b +/- sqrt(b²-4ac)]/2a

x = [-7 +/- sqrt(7² - 4(1)(-4)]/2

x = [-7 +/- sqrt(65)]/2

Answer 2

Answer:

B

Step-by-step explanation:

Our equation is x² + 7x - 4 = 0. We need to use the quadratic formula, which states that for a quadratic formula of the form ax² + bx + c = 0, then the zeroes/solutions are: $x=\frac{-b+\sqrt{b^2-4ac} }{2a}$ or $x=\frac{-b-\sqrt{b^2-4ac} }{2a}$.

Here, a = 1, b = 7, and c = -4. Plug these in:

$x=\frac{-b+\sqrt{b^2-4ac} }{2a}$

$x=\frac{-7+\sqrt{7^2-4*1*(-4)} }{2*1}=\frac{-7+\sqrt{49+16} }{2} =\frac{-7+\sqrt{65} }{2}$

OR

$x=\frac{-b-\sqrt{b^2-4ac} }{2a}$

$x=\frac{-7-\sqrt{7^2-4*1*(-4)} }{2*1}=\frac{-7-\sqrt{49+16} }{2} =\frac{-7-\sqrt{65} }{2}$

Thus, the answer is B.