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Answers
Answer:
For the roots of a quadratic equation in general form, ax2+bx+c=0, to be equal, the quadratic formula shows that the discriminant, b2−4ac, must equal zero.
Thus, (n−l)2−4(m−n)(l−m)=0
⟹(n2−2ln+l2)−4(lm−m2−ln+mn)=0
⟹n2−2ln+l2−4lm+4m2+4ln−4mn=0
⟹l2+4m2+n2−4lm+2ln−4mn=0
⟹(l+n−2m)2=0
⟹l+n−2m=0
Now, to be an AP, the difference between l and m must equal the difference between m and n:
m−l=n−m⟹l+n−2m=0
QED
(Thanks to for the error-check that let me complete my answer! :)
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The question needs a little bit of editing. I will correct the question before answering it.
If the roots of the quadratic equation (m−n)x2+(n−ℓ)x+(ℓ−m)=0 are equal, then show that ℓ, m, n are in AP.
A necessary and sufficient condition for the roots to be equal is
(n−ℓ)2=4(m−n)(ℓ−m).
Writing n−ℓ as (n−m)+(m−ℓ) and squaring gives
(m−n)2+2(m−n)(ℓ−m)+(ℓ−m)2=4(m−n)(ℓ−m).
Therefore
0=((m−n)−(ℓ−m))2=(2m−(ℓ+n))2,
so that 2m=ℓ+n. Hence ℓ, m, n are in AP. ■
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Answer:
mn(x²+1) = (m²+n²)x
=> mnx² + mn - (m²+n²)x = 0
=> mnx² - m²x - n²x + mn = 0
=> mx(nx-m) -n(nx-m) = 0
=> (mx-n) *(nx-m) = 0
=> x = n/m or m/n