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6.已知一元二次方程$x^2 + 4x - 3 = 0$的两根是$x_1,x_2$，则$x_1^2 + x_2^2$的值为
22
。答案
6. 22
7.已知$m,n$是关于$x$的一元二次方程$x^2 - 2025x + 1 = 0$的两个根，则$(m+1)(n+1) =$
2027
.答案
7. 2027
8.若关于$x$的方程$x^2 - 2(m+1)x + m + 4 = 0$的两根的倒数和为1，则$m$的值为
2
.答案
8. 2
9.(2024·四川遂宁中考)已知关于$x$的一元二次方程$x^2-(m+2)x+m-1=0$.
(1)求证:无论$m$取何值,方程都有两个不相等的实数根;
(2)如果方程的两个实数根为$x_1,x_2$,且$x_1^2+x_2^2 -x_1x_2=9$,求$m$的值.
(1)求证:无论$m$取何值,方程都有两个不相等的实数根;
(2)如果方程的两个实数根为$x_1,x_2$,且$x_1^2+x_2^2 -x_1x_2=9$,求$m$的值.
答案
9.(1)证明:$x^2-(m+2)x+m-1=0$,
这里$a=1,b=-(m+2),c=m-1$,
$\Delta =b^2-4ac$
$=[-(m+2)]^2-4×1×(m-1)$
$=m^2+4m+4-4m+4$
$=m^2+8$.
$\because m^2≥0$,
$\therefore \Delta＞0$.
$\therefore$无论$m$取何值,方程都有两个不相等的实数根.
(2)解:设方程$x^2-(m+2)x+m-1=0$的两个实数根为$x_1,x_2$,
则$x_1+x_2=m+2,x_1x_2=m-1$.
$\because x_1^2+x_2^2-x_1x_2=9$,即$(x_1+x_2)^2-3x_1x_2=9$,
$\therefore (m+2)^2-3(m-1)=9$,
整理得$m^2+m-2=0$.
$\therefore (m+2)(m-1)=0$,
解得$m_1=-2,m_2=1$.
$\therefore m$的值为$-2$或$1$.
这里$a=1,b=-(m+2),c=m-1$,
$\Delta =b^2-4ac$
$=[-(m+2)]^2-4×1×(m-1)$
$=m^2+4m+4-4m+4$
$=m^2+8$.
$\because m^2≥0$,
$\therefore \Delta＞0$.
$\therefore$无论$m$取何值,方程都有两个不相等的实数根.
(2)解:设方程$x^2-(m+2)x+m-1=0$的两个实数根为$x_1,x_2$,
则$x_1+x_2=m+2,x_1x_2=m-1$.
$\because x_1^2+x_2^2-x_1x_2=9$,即$(x_1+x_2)^2-3x_1x_2=9$,
$\therefore (m+2)^2-3(m-1)=9$,
整理得$m^2+m-2=0$.
$\therefore (m+2)(m-1)=0$,
解得$m_1=-2,m_2=1$.
$\therefore m$的值为$-2$或$1$.
10.(2025·四川南充中考)设$x_1,x_2$是关于$x$的方程$(x-1)(x-2)=m^2$的两根.
(1)当$x_1=-1$时,求$x_2$及$m$的值;
(2)求证:$(x_1-1)(x_2-1)≤0$.
(1)当$x_1=-1$时,求$x_2$及$m$的值;
(2)求证:$(x_1-1)(x_2-1)≤0$.
答案
10.(1)解:把$x_1=-1$代入方程$(x-1)(x-2)=m^2$,
得$m^2=6,\therefore m=\pm\sqrt{6}$.
$\therefore (x-1)(x-2)=6$,
即$x^2-3x-4=0$.
$\therefore (x-4)(x+1)=0$.
$\therefore x_1=-1,x_2=4$.
$\therefore x_2=4,m=\pm\sqrt{6}$.
(2)证明:方程$(x-1)(x-2)=m^2$可化为$x^2-3x+2-m^2=0$.
$\because \Delta=9-4(2-m^2)=4m^2+1＞0$,
$\therefore$方程有两个不相等的实数根.
$\because$方程$(x-1)(x-2)=m^2$即$x^2-3x+2-m^2=0$的两根为$x_1,x_2$,
$\therefore x_1+x_2=3,x_1· x_2=2-m^2$.
$\therefore (x_1-1)(x_2-1)=x_1· x_2-(x_1+x_2)+1=2-m^2-3+1=-m^2$.
$\because m^2≥0$,
$\therefore -m^2≤0$,即$(x_1-1)(x_2-1)≤0$.
得$m^2=6,\therefore m=\pm\sqrt{6}$.
$\therefore (x-1)(x-2)=6$,
即$x^2-3x-4=0$.
$\therefore (x-4)(x+1)=0$.
$\therefore x_1=-1,x_2=4$.
$\therefore x_2=4,m=\pm\sqrt{6}$.
(2)证明:方程$(x-1)(x-2)=m^2$可化为$x^2-3x+2-m^2=0$.
$\because \Delta=9-4(2-m^2)=4m^2+1＞0$,
$\therefore$方程有两个不相等的实数根.
$\because$方程$(x-1)(x-2)=m^2$即$x^2-3x+2-m^2=0$的两根为$x_1,x_2$,
$\therefore x_1+x_2=3,x_1· x_2=2-m^2$.
$\therefore (x_1-1)(x_2-1)=x_1· x_2-(x_1+x_2)+1=2-m^2-3+1=-m^2$.
$\because m^2≥0$,
$\therefore -m^2≤0$,即$(x_1-1)(x_2-1)≤0$.
