If the roots of equation (p^2+q^2)x^2-2q(p+r)x+(p^2+r^2) = 0 are equal. The value of q^2 is
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Step-by-step explanation:
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If the roots of the quadratic equation: p(q-r)x^2+q(r-p)x+r(p-q)=0 are equal, show that:1/p + 1/r = 2/q
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jainparul1973
jainparul1973 Ambitious
First find the discriminant of the given quadratic equation by formula (b^2 -4ac)
also d=0
so the equation will come.
then solve the equation...
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mysticd
mysticd Genius
Answer:
\green {\frac{1}{p}+\frac{1}{r}=\frac{2}{q}}
Step-by-step explanation:
Given:
The roots of the quadratic equation :
p(q-r)x²+q(r-p)x+r(p-q)=0 are equal.
To prove :
\frac{1}{p}+\frac{1}{r}=\frac{2}{q}
Solution:
Compare given Quadratic equation with ax²+bx+c=0, we get
a = p(q-r), b = q(r-p), c = r(p-q)
Discreminant (D) = 0
/* roots are equal given */
=> b²-4ac=0
=>[q(r-p)]²-4×p(q-r)×r(p-q)=0
=>(qr-pq)²-4pr(q-r)(p-q)=0
=> (qr)²+(pq)²-4(qr)(pq)-4pr(pq-q²-pr+qr)=0
=> (qr)²+(pq)²-4pq²r-4p²qr+4prq²+4p²r²-4pqr²=0
=> (qr)²+(pq)²+(-2pr)²+2pq²r-4p²qr-4pqr²=0
=> (qr)²+(pq)²+(-2pr)²+2(qr)(pq)+2(pq)(-2pr)+2(-2pr)(qr)=0
/* we know the algebraic identity*/
/*a²+b²+c²+2ab+2bc+2ca=(a+b+c)² */
=> (qr+pq-2pr)² = 0
=> qr+pq-2pr = 0
Divide each term by pqr , we get
\implies\frac{qr}{pqr}+\frac{pq}{pqr}-\frac{2pr}{pqr}=0
\implies \frac{1}{p}+\frac{1}{r}-\frac{2}{q}=0
Therefore,
\frac{1}{p}+\frac{1}{r}=\frac{2}{q}