7√x-2x+15=0

To solve the equation \( 7\sqrt{x} - 2x + 15 = 0 \), let's follow these steps: 1. **Isolate the square root term**: \[ 7\sqrt{x} = 2x - 15 \] 2. **Square both sides to eliminate the square root** (be cautious as this can introduce extraneous solutions): \[ (7\sqrt{x})^2 = (2x - 15)^2 \] \[ 49x = 4x^2 - 60x + 225 \] 3. **Rearrange the equation** into standard quadratic form: \[ 0 = 4x^2 - 60x + 225 - 49x \] \[ 0 = 4x^2 - 109x + 225 \] 4. **Use the quadratic formula** to solve for \( x \): The quadratic formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 4 \), \( b = -109 \), and \( c = 225 \). 5. **Calculate the discriminant**: \[ b^2 - 4ac = (-109)^2 - 4 \cdot 4 \cdot 225 \] \[ = 11881 - 3600 \] \[ = 8281 \] 6. **Calculate the square root of the discriminant**: \[ \sqrt{8281} = 91 \] 7. **Substitute back into the quadratic formula**: \[ x = \frac{109 \pm 91}{8} \] This gives us two potential solutions: \[ x_1 = \frac{200}{8} = 25 \] \[ x_2 = \frac{18}{8} = 2.25 \] 8. **Check for extraneous solutions** by substituting back into the original equation: - For \( x = 25 \): \[ 7\sqrt{25} - 2(25) + 15 = 35 - 50 + 15 = 0 \quad \text{(Valid)} \] - For \( x = 2.25 \): \[ 7\sqrt{2.25} - 2(2.25) + 15 = 7(1.5) - 4.5 + 15 = 10.5 - 4.5 + 15 = 21 \quad \text{(Not valid)} \] Thus, the only solution to the equation \( 7\sqrt{x} - 2x + 15 = 0 \) is: \[ \boxed{25} \]