初三难度一元二次方程题目 1. 已知关于x的一元二次方程(m - 1)x^{2}+2mx + m+3=0有两个不相等的实数根,求m的取值范围。 2. 设关于x的一元二次方程x^{2}+2ax+b^{2}=0,若是从0,1,2,3四个数中任取... 初三难度一元二次方程题目 1. 已知关于x的一元二次方程(m - 1)x^{2}+2mx + m+3=0有两个不相等的实数根,求m的取值范围。 2. 设关于x的一元二次方程x^{2}+2ax+b^{2}=0,若是从0,1,2,3四个数中任取的一个数,b是从0,-1,2三个数中任取的一个数,求上述方程有实根的概率。 3. 某商场销售一批名牌衬衫,平均每天可售出20件,每件盈利40元,为了扩大销售,增加盈利,尽快减少库存,商场决定采取适当的降价措施,经调查发现,如果每件衬衫每降价1元,商场平均每天可多售出2件,若商场平均每天要盈利1200元,每件衬衫应降价多少元? 4. 已知x_1,x_2是一元二次方程x^{2}-4x+1=0的两个实数根,求(x_1+x_2)^2\div(\frac{1}{x_1}+\frac{1}{x_2})的值。 5. 对于一元二次方程ax^{2}+bx+c=0(a\neq0),已知a-b+c=0,且4a+2b+c=0,求该方程的两个根 Read more

Steps to Solve

Here's a breakdown of each problem with steps to solve:

1. Finding the range of m for distinct real roots For the quadratic equation $(m - 1)x^2 + 2mx + m + 3 = 0$ to have two distinct real roots, two conditions must be met: 1) The coefficient of the quadratic term must not be zero, i.e., $m - 1 \neq 0$, and 2) the discriminant must be greater than zero, i.e., $\Delta > 0$. The discriminant $\Delta$ is given by $\Delta = b^2 - 4ac$, where $a = m - 1$, $b = 2m$, and $c = m + 3$.

2. Calculate the discriminant and apply the condition Calculate the discriminant: $\Delta = (2m)^2 - 4(m - 1)(m + 3) = 4m^2 - 4(m^2 + 3m - m - 3) = 4m^2 - 4(m^2 + 2m - 3) = 4m^2 - 4m^2 - 8m + 12 = -8m + 12$ For distinct real roots, $\Delta > 0$, so $-8m + 12 > 0$.

3. Solve the inequality for m $-8m > -12$ $m < \frac{-12}{-8}$ $m < \frac{3}{2}$

4. Consider the condition $m-1 \neq 0$ $m \neq 1$

5. Combine the conditions Combining $m < \frac{3}{2}$ and $m \neq 1$, the range for $m$ is $m < \frac{3}{2}$ and $m \neq 1$.

6. Probability of real roots The quadratic equation is $x^2 + 2ax + b^2 = 0$. For real roots, the discriminant $\Delta = (2a)^2 - 4(1)(b^2) \geq 0$. Thus $4a^2 - 4b^2 \geq 0$, which simplifies to $a^2 \geq b^2$ or $|a| \geq |b|$.

7. List possible values of a and b $a$ can be $0, 1, 2, 3$ and $b$ can be $0, -1, 2$. Total number of pairs $(a, b)$ is $4 \times 3 = 12$.

8. Find pairs that satisfy $|a| \geq |b|$ The pairs that satisfy $|a| \geq |b|$ are:

   • $a = 0$: $b = 0$
   • $a = 1$: $b = 0, -1$
   • $a = 2$: $b = 0, -1, 2$
   • $a = 3$: $b = 0, -1, 2$ So the number of pairs that satisfy the condition is $1 + 2 + 3 + 3 = 9$.

9. Calculate the probability The probability is $\frac{9}{12} = \frac{3}{4}$.

10. Profit maximization problem Let $x$ be the amount the price is reduced per shirt. The number of shirts sold is $20 + 2x$. The profit per shirt is $40 - x$. The total profit is $(20 + 2x)(40 - x)$. We want this to be equal to 1200, so $(20 + 2x)(40 - x) = 1200$.

11. Solve the equation for $x$ $800 - 20x + 80x - 2x^2 = 1200$ $-2x^2 + 60x - 400 = 0$ $x^2 - 30x + 200 = 0$ $(x - 10)(x - 20) = 0$ $x = 10$ or $x = 20$

12. Determine the price reduction If $x = 10$, the price is reduced by 10, and the profit per shirt is 30, and the number of shirts sold is 40. If $x = 20$, the price is reduced by 20, and the profit per shirt is 20, and the number of shirts sold is 60. Both values yield a total profit of 1200. The problem asks for the price reduction, so in the absence of other guidance, either answer is acceptable. I'll choose the minimum price reduction.

13. Using Vieta's formulas and simplification For the equation $x^2 - 4x + 1 = 0$, by Vieta's formulas, $x_1 + x_2 = 4$ and $x_1x_2 = 1$. We want to find $\frac{(x_1 + x_2)^2}{\frac{1}{x_1} + \frac{1}{x_2}} = \frac{(x_1 + x_2)^2}{\frac{x_1 + x_2}{x_1x_2}} = \frac{(x_1 + x_2)^2 x_1x_2}{x_1 + x_2} = (x_1 + x_2)x_1x_2$.

14. Substitute Vieta's formulas $(x_1 + x_2)x_1x_2 = (4)(1) = 4$.

15. Solving for the roots Given $ax^2 + bx + c = 0$ with $a - b + c = 0$ and $4a + 2b + c = 0$. Since $a - b + c = 0$, $x = -1$ is a root. Also, from $a - b + c = 0$, we get $c = b - a$. Substituting this into $4a + 2b + c = 0$, we get $4a + 2b + b - a = 0$, so $3a + 3b = 0$, implying $b = -a$. Then $c = b - a = -a - a = -2a$. Thus, the quadratic equation is $ax^2 - ax - 2a = 0$. Since $a \neq 0$, we can divide by $a$, yielding $x^2 - x - 2 = 0$.

16. Factor the quadratic equation Factoring gives $(x - 2)(x + 1) = 0$, so $x = 2$ or $x = -1$.