The relationship between commutators and functions and equations is used to solve more complex univariate quadratic equation problem analysis and solution applications

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<h1 class="pgc-h-arrow-right" data-track="2" > analysis and solution</h1>

Both equations contain three parameters a, b, and c, and the direct solution is more complicated, but the two look very similar, there must be a certain relationship, as long as the internal relationship between the two can be found, the problem will be solved.

1. Use the commutation method to solve the problem

X in the first equation is equivalent to x-2 in the second equation, and equation (2) makes m=x-2, then equation (2) becomes

a(m+c)^2+b=0, the same as equation (1), so the value of m is -2 or 1

When m=-2, i.e. x-2=-2, x=0;

When m=1, i.e. x-2=1, x=3.

Therefore, the two roots of equation (2) are 0,3.

2. Use the relationship between functions and equations to solve problems

The solution of equation (1) can be seen as the abscissa of the quadratic function y=a(x+c)^2+b at the intersection of the x axis, and the solution of equation (2) can be seen as the abscissa of the quadratic function y=a(x+c-2)^2+b at the intersection of the x-axis. The second function can be obtained by translating the first function by 2 units to the right, and the solution of the corresponding equation should also add 2, and the two roots of equation (1) are -2, 1, so the two roots of equation (2) are 0 and 3.

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In the previous article, the most complex univariate quadratic equations 2 (x+1)^2-3=0 and 2(x+1)^2+3(x+1)^2+3(x+1)=0 can be easily obtained from simple unary quadratic equations 2x^2-3=0 and 2x^2+3x=0.

For more complex unary quadratic equations, there are often simple algorithms, pay attention to observation, and find out the rules from them. The so-called first use the brain before starting to use, and try to simplify as much as possible.