Find the roots CSM4x²-23x=-28​

To find the roots of the equation \(4x^2 - 23x + 28 = 0\), we can use the quadratic formula, which is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]In this equation, \(a = 4\), \(b = -23\), and \(c = 28\).1. First, calculate the discriminant \(b^2 - 4ac\):\[b^2 = (-23)^2 = 529\]\[4ac = 4 \cdot 4 \cdot 28 = 448\]\[b^2 - 4ac = 529 - 448 = 81\]2. Since the discriminant is positive, we have two real roots.3. Now, substitute the values into the quadratic formula:\[x = \frac{-(-23) \pm \sqrt{81}}{2 \cdot 4}\]\[x = \frac{23 \pm 9}{8}\]4. Now, calculate the two roots:For the first root:\[x_1 = \frac{23 + 9}{8} = \frac{32}{8} = 4\]For the second root:\[x_2 = \frac{23 - 9}{8} = \frac{14}{8} = \frac{7}{4}\]So, the roots of the equation \(4x^2 - 23x + 28 = 0\) are:\[x = 4 \74

Answered by romnickpallon | 2024-09-05