(iv) $(x-3)(2 x+1)=x(x+5)$
Answer (1)
Expand both sides of the equation: ( x − 3 ) ( 2 x + 1 ) = x ( x + 5 ) becomes 2 x 2 − 5 x − 3 = x 2 + 5 x .
Simplify to the standard quadratic form: x 2 − 10 x − 3 = 0 .
Apply the quadratic formula: x = 2 a − b ± b 2 − 4 a c .
Obtain the solutions: x = 5 ± 2 7 .
Explanation
Expanding the equation First, let's expand both sides of the equation to get rid of the parentheses. This will help us simplify the equation and bring all the terms to one side.
Expanding the left side Expanding the left side, we have: ( x − 3 ) ( 2 x + 1 ) = 2 x 2 + x − 6 x − 3 = 2 x 2 − 5 x − 3
Expanding the right side Expanding the right side, we get: x ( x + 5 ) = x 2 + 5 x
Equating both sides Now, let's set the expanded forms equal to each other: 2 x 2 − 5 x − 3 = x 2 + 5 x
Rearranging the equation Next, we want to move all the terms to one side to set the equation to zero. Subtract x 2 and 5 x from both sides: 2 x 2 − x 2 − 5 x − 5 x − 3 = 0
Simplifying to quadratic form Combining like terms, we obtain a quadratic equation: x 2 − 10 x − 3 = 0
Applying the quadratic formula Now, we can use the quadratic formula to solve for x . The quadratic formula is given by: x = 2 a − b ± b 2 − 4 a c
Substituting values In our equation, a = 1 , b = − 10 , and c = − 3 . Plugging these values into the quadratic formula, we get: x = 2 ( 1 ) − ( − 10 ) ± ( − 10 ) 2 − 4 ( 1 ) ( − 3 )
Simplifying the square root Simplifying further: x = 2 10 ± 100 + 12 x = 2 10 ± 112
Simplifying the radical We can simplify 112 as 16 ⋅ 7 = 4 7 . So, x = 2 10 ± 4 7
Final solutions Finally, we can divide both terms in the numerator by 2: x = 5 ± 2 7
Examples
Understanding how to solve quadratic equations is crucial in many fields, such as physics and engineering. For example, when calculating the trajectory of a projectile, you often need to solve a quadratic equation to find the time it takes for the projectile to reach a certain height or distance. Similarly, in engineering, quadratic equations are used to design circuits, calculate stresses in materials, and optimize various systems. Mastering these equations provides a strong foundation for tackling real-world problems in these disciplines. Let's say you're designing a bridge and need to calculate the parabolic curve of an arch. The equation might look like y = a x 2 + b x + c , and solving for 'x' at a specific height 'y' involves solving a quadratic equation.
