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If 'p'and 'q' are the zeros of the quadratic polynomial f(x)= x²-x-4,then find the value of 1/p+1/q-pq |
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Answer» Answer: The second form won’t be INDEPENDENT of which root is p and which is q, so probably a typo. It would lead to 2 different results including complex coefficients. (x-p)*(x-q)=x^2-x*(p+q)+p*q=x^2–2*x+3 so p+q=2, p*q=3 ROOTS p+2,q+2 so (x-p-2)*(x-q-2)=x^2+x*(-q-2-p-2)+(p+2)*(q+2)= x^2+(-p-q-4)*x+(p*q+2*p+2*q+4). SINCE p+q=2 and p*q=3, we GET x^2+(-4–2)*x+(3+2*2+4)=x^2–6*x+11 For second form, use (p-1)/(p+1) and (q-1)/(q+1) first, to get (x-(p-1)/(p+1))*(x-(q-1)/(q+1))=x^2+x*(-(p-1)/(p+1)-(q-1)/(q+1))+(p-1)*(q-1)/((p+1)*(q+1))=x^2+x*(2–2*p*q)/(p*q+p+q+1)+(p*q-p-q+1)/(p*q+p+q+1)= x^2+x*(2–2*3)/(3+2+1)+(3–2+1)/(3+2+1)=x^2–2*x/3+1/3 so 3*x^2–2*x+1 and secondly, (p+1)/(p-1) and (q+1)/(q-1) gives x^2–2*x+3 which is the original equation, since the roots are simply re-ordered. Step-by-step explanation: I HOPE it helps you |
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