不知樓主的題目是否為：
已知二次函數(shù)y=x²-(m²+8)x+2(m²+6).設(shè)這個(gè)函數(shù)圖像與x軸交于B,C兩點(diǎn),與y軸交于點(diǎn)A,若△ABC的面積為48,求m的值?
如果二次函數(shù)是上式,則解答如下.
由函數(shù)與X軸交于B、C兩點(diǎn)得
判別式△>0,
[-(m²+8)]²-4*2(m²+6)>0
得(m²+4)>0,
m為任意實(shí)數(shù)R.
由已知得|BC|*|AO|*1/2=48··········(1)
當(dāng)x=0時(shí),函數(shù)與y軸線交點(diǎn)坐標(biāo)為(0,2(m²+6))
則得|AO|=2(m²+6)··········(2)；
令B、C兩點(diǎn)坐標(biāo)分別為(x1,0)、(x2,0)
則得|BC|=|x2-x1|,
由根與系數(shù)的關(guān)系得
x1+x2=m²+8,
x1*x2=2(m²+6)
對(duì)|BC|=|x2-x1|兩邊平方得
|BC|²=|x2-x1|²
=(x2-x1)²
=x2²+x1²-2x1*x2
=(x1+x2)²-4x1*x2············(3)
將x1+x2=m²+8,x1*x2=2(m²+6)代入(3)式得
|BC|²=(m²+8)²-4*2(m²+6)
=m²+4···········(4)
將以上(2)與(4)式代入(1)式,得
(m²+4)*2(m²+6)*1/2=48
(m²+4)*(m²+6)=48
m^4+10m²-24=0
(m²-2)(m²+12)=0
解得
m²=2,m²=-12(舍去)
即得m=±√2.