Question

7x^2=9+x what are the values of xWill give medal and points

Answer 1

7x² = 9 + x   Subtract x from both sides
7x² - x = 9    Subtract 9 from both sides
7x² - x - 9 = 0   Use the Quadratic Formula

a = 7 , b = -1 , c = -9

x = $\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$   Plug in the a, b, and c values
x = $\frac{- (-1) \pm \sqrt{(-1)^2 - 4(7)(-9)} }{2(7)}$   Cancel out the double negative
x = $\frac{1 \pm \sqrt{(-1)^2 - 4(7)(-9)} }{2(7)}$   Square -1
x = $\frac{1 \pm \sqrt{1 - 4(7)(-9)} }{2(7)}$   Multiply 7 and -9
x = $\frac{1 \pm \sqrt{1 - 4(-63} }{2(7)}$   Multiply -4 and -63
x = $\frac{1 \pm \sqrt{1 + 252} }{2(7)}$   Multiply 2 and 7
x = $\frac{1 \pm \sqrt{1 + 252} }{14}$   Add 1 and 252
x = $\frac{1 \pm \sqrt{253} }{14}$   Split up the $\pm$
x = $\left \{ {{ \frac{1 + \sqrt{253} }{14} } \atop { \frac{1 - \sqrt{253} }{14} }} \right.$
The approximate square root of 253 is 15.905973.
x ≈ $\left \{ { \frac{1 + 15.905973}{14} } \atop { \frac{1 - 15.905973}{14} }} \right$   Add and subtract
x ≈ $\left \{ {{ \frac{16.905973}{14} } \atop { \frac{14.905973}{14} }} \right.$   Divide
x ≈ $\left \{ {{1.2075} \atop {1.0647}} \right.$   Round to the nearest hundredth
x ≈ $\left \{ {{1.21} \atop {1.06}} \right.$