The solution to [tex]x^2-10 x=24[/tex] is $\square$.
The solution to [tex]2 x^2-11=87[/tex] is $\square$.
The solution to [tex]3 x^2-12 x+24=0[/tex] is $\square$.
Answer (2)
Solve x 2 − 10 x = 24 by factoring: x 2 − 10 x − 24 = ( x − 12 ) ( x + 2 ) = 0 , so x = 12 or x = − 2 .
Solve 2 x 2 − 11 = 87 by isolating x 2 : 2 x 2 = 98 , x 2 = 49 , so x = ± 7 .
Solve 3 x 2 − 12 x + 24 = 0 using the quadratic formula after dividing by 3: x 2 − 4 x + 8 = 0 , so x = 2 4 ± ( − 4 ) 2 − 4 ( 1 ) ( 8 ) = 2 ± 2 i .
The solutions are 12 , − 2 ; 7 , − 7 ; 2 + 2 i , 2 − 2 i .
Explanation
Problem Analysis We are given three quadratic equations to solve:
x 2 − 10 x = 24
2 x 2 − 11 = 87
3 x 2 − 12 x + 24 = 0
Our objective is to find the solutions to these equations.
Solving the First Equation Let's solve the first quadratic equation: x 2 − 10 x = 24 . We can rewrite this as x 2 − 10 x − 24 = 0 . We need to find two numbers that multiply to -24 and add up to -10. These numbers are -12 and 2. Therefore, we can factor the quadratic as ( x − 12 ) ( x + 2 ) = 0 . This gives us two possible solutions: x = 12 or x = − 2 .
Solving the Second Equation Now, let's solve the second quadratic equation: 2 x 2 − 11 = 87 . We can rewrite this as 2 x 2 = 98 . Dividing both sides by 2, we get x 2 = 49 . Taking the square root of both sides, we find x = ± 7 . So the solutions are x = 7 and x = − 7 .
Solving the Third Equation Finally, let's solve the third quadratic equation: 3 x 2 − 12 x + 24 = 0 . We can simplify this equation by dividing all terms by 3, which gives us x 2 − 4 x + 8 = 0 . Since this quadratic does not factor easily, we will use the quadratic formula: x = 2 a − b ± b 2 − 4 a c In this case, a = 1 , b = − 4 , and c = 8 . Plugging these values into the quadratic formula, we get: x = 2 ( 1 ) 4 ± ( − 4 ) 2 − 4 ( 1 ) ( 8 ) = 2 4 ± 16 − 32 = 2 4 ± − 16 = 2 4 ± 4 i = 2 ± 2 i So the solutions are x = 2 + 2 i and x = 2 − 2 i .
Final Answer Therefore, the solutions to the three quadratic equations are:
x 2 − 10 x = 24 has solutions x = 12 and x = − 2 .
2 x 2 − 11 = 87 has solutions x = 7 and x = − 7 .
3 x 2 − 12 x + 24 = 0 has solutions x = 2 + 2 i and x = 2 − 2 i .
Examples
Quadratic equations are not just abstract math; they appear in various real-world applications. For instance, when designing a bridge, engineers use quadratic equations to calculate the curve of an arch or the trajectory of a cable. Similarly, in physics, these equations help determine the path of a projectile, like a ball thrown in the air. Understanding how to solve quadratic equations is crucial for making accurate predictions and ensuring structural integrity in engineering and physics.
The solutions to the equations are as follows: x 2 − 10 x = 24 gives x = 12 and x = − 2 ; 2 x 2 − 11 = 87 gives x = 7 and x = − 7 ; 3 x 2 − 12 x + 24 = 0 gives x = 2 + 2 i and x = 2 − 2 i .
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