A quadratic equation error-prone point

A quadratic equation error-prone point

Outline

In the examination of a quadratic equation is simple in mathematical level, but because of its limited scope, there are many error-prone points.

Error-prone point

1. decision

Requirements: ① unknown ② ③ maximum secondary Zhengshi equation (the number of times before, fractional or simple look after radical simplification of Zhengshi see equation)

The request parameter value times the equation
on \ (X \) equation \ (\ displaystyle (m + 1 ) x ^ {m ^ 2 + 1} + (m-3) x-1 = 0 \) a linear equation, the [ \ (m = \)   \ (0 or -1 \)   ]

Order term for the variable to be considered absent (coefficient of 0) and a 1 or 0 in the case of secondary

2. solving equations

note

1. Do not write the equation has no solution,
2. Quadratic equation can only be two solutions without real roots of the equation must be written in one line (line feed "and" can not be represented or relationship)

Kaiping direct method

\ ((3x-5) ^
2 = a \) when \ (a <0 \) when no real roots of the equation
as \ (a = 0 \) , the \ (3x-. 5 = 0 \) (must be individually 0 different considerations, calculated results and prescribing)
\ (\ DisplayStyle x_1 = x_2 = \ {FRAC. 3. 5} {} \) (Chemical equation time still has two solutions)
when \ (a> 0 \ ) , the ............

With methods

Formula Description \ (\ neq0 \)
to prove certain \ (> 0 \) or \ (<0 \) , and then start a new line saying \ (\ neq0 \)

Official method

Root Formula
When \ (b ^ 2-4ac \ geq0 \ ) when, \ (\ DisplayStyle X = \ {FRAC -b \ PM \ 2-4ac sqrt {B}} ^ {2} \)

When \ (b ^ 2-4ac <0 \ ) , then the equation has no real roots

The whole law

Known \ (X ^ 2-Y ^ 2 + XY = 0 \) , seeking \ (\ displaystyle \ frac {x } {y} (y \ neq0) \)

The \ (\ displaystyle \ frac {x } {y} \) seen as a whole
\ (\ DisplayStyle \ FRAC {2} X ^ Y ^ 2} + {\ X FRAC {Y} {-1} = 0 \)
\ (\ displaystyle \ frac {x}
{y} = \ frac {-1 \ pm \ sqrt {5}} {2} \) a \ (Y \) considered as critical parameters operator
\ (\ displaystyle x = \ frac { -Y \ PM \ sqrt {Y}. 5} {2} \)
\ (\ DisplayStyle ∴ \ FRAC {X} {Y} = \ FRAC {-1 + \. 5 sqrt {2}}} {\)

After using the whole law (but not limit the subject domain), on behalf of the return to test

Known \ (\ DisplayStyle X ^ 2 + \ FRAC {. 1} {X ^ 2} + X + \ FRAC {. 1} {X} = 0 \) , seeking \ (\ displaystyle x + \ frac {1} {x} \ ) values

\ (\ DisplayStyle (X + \ FRAC {. 1} {X}) ^ 2+ (X + \ FRAC {. 1} {X}) - 2 = 0 \)
\ (\ DisplayStyle X + \ FRAC {. 1} {X} = ... ... = \ frac {-1 \ pm3
} {2} \) when \ (\ displaystyle x + \ frac {1} {x} = - 2 \) when, ......
\ (x_1 = x_2 = -1 \)
when \ (\ displaystyle x + \ frac { 1} {x} = 1 \) when, ......
\ (^ 2-4ac B = (-. 1) ^ 2-4 <0 \) has no real roots, rounding
summary, ... ... \ (2- ^ X | X | -2 = 0 \)
\ (| X | ^ 2- | X | -2 = 0 \)
\ ((| X | -2) (| X | + 1'd) = 0 \) \ (| X | taken 2 \)
\ (2 = x_1, x_2 = -2 \)

Root of the discussion

Be sure to consider \ (a = 0 \) case, Category talk

Vieta theorem

content

For \ (^ AX + BX + C 2 = 0 (A \ neq0) \) , in (b ^ 2-4ac \ geq0 \) \ when there \ (\ displaystyle x_1 + x_2 = - \ frac {b} { a}, x_1x_2 = \ frac { c} {a} \)

When the equation coefficients letters, to sentence the root , all numerical coefficients do not

About known \ (X \) equation \ (x ^ {2} + (2 k-1) x + k ^ {2} -1 = 0 \) square and two for the \ (9 \) , seeking \ (K \) values

①
\[b^2-4ac=(2k-1)^2-4(k^2-1)=-4k+5\geq 0\]
\[k\leq\frac{5}{4}\]
②
\[x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2 \]
\[=(2k-1)^2-2(k^2-1)\]
\[=4k^2-4k+1-2k^2+2\]
\[=2k^2-4k+3=9\]
\[2k^2-4k-6=0\]
\[(k+1)(k-3)=0\]
\[k_1=-1,k_2=3>\frac54(\text{舍})\]

"Root" (×× ×× is the root)

Substitution method

The whole law

With a large dividing the algebraic known to remove (and then replace the constant \ (\ Times \) product of another algebraic) containing the partial factorization letters, into the product of formula or known to contain the expression 0 (with some difficulty thinking)

Vieta theorem

Premise: homogeneous (missing first times down times)

It can be the original equation by substituting the portion of the drop-segment order equation, followed by a state to form, and then can be factorized + Vieta theorem

application

Column according to casual working equation

All figures have to pay attention to the casual working in the actual situation, we must go back generations, casual working conditions, pay attention to the implicit ( "Minimalist secondary radical", be sure to test the simplest)

Sales issues

Computationally expensive, there is a certain degree of difficulty Note purposes described in casual working "in order to reduce inventory as soon as possible" "To enable customers to enjoy the benefits."

Nanjing students less able to calculate, soOral regulationsAll Big Data equation can be multiplied by the cross, do not do the math Wuxi, they have a calculator.