An Insight into Quadratic Equations and Cubic Equations with Real Coefficients

Transcription

1 An Insight into Qurti Equtions n Cubi Equtions with Rel Coeffiients Qurti Equtions A qurti eqution is n eqution of the form x + bx + =, where o It n be solve quikly if we n ftorize the expression x + bx + If we fin x + bx + = x ( α)( x β ), then x ( α)( x β ) = les to the solutions x= α, or x= β In the se of enountering some iffiulty in the ftoriztion, we hve the formul b± b x = The formul ensures the solution of the eqution, if there is ny, or otherwise revels its sitution. The importnt omponent in the formul is the isriminnt = b Reltions between roots n oeffiients of qurti eqution Assuming the roots of x + bx + = to be α, β We n re-write the eqution s x ( α)( x β ) = It is ler then tht x + bx + ( x α)( x β ) x [ ( α + β) x+ αβ] b So α + β = n αβ = These results n lso be eue by equting α, β to b+ b b b, respetively n then work out α + β n αβ If there is no rel root for the eqution, the expression f ( x) = x + bx+ nnot hnge sign s we vry the vlues of x through ll numbers, tht is, it must remin either positive or negtive throughout, otherwise oring to ontinuity, for f ( x) to hnge from positive vlue to negtive vlue, it must be zero somewhere, sy t x = α, n tht is the root of f( x ) =. Hene f ( x) = x + bx+ is either positive or negtive for ll vlues of x if the isriminnt = b <, tht is, when the eqution f( x ) = hs no rel root. Cubi Eqution A ubi eqution is n eqution of the form x + bx + x + =, where o 1

2 Let g( x) = x + bx + x+ While we sy tht qurti (rel) eqution my not hve rel root, ubi (rel) eqution hs t lest one rel root. For if L is lrge vlue, the sign of g( L ) is ifferent from the sign of g( L) (Full isussion is tken s n exerise) So, given ubi (rel) eqution, it is wise to fin by tril one root first, for there is ertinly one With one root foun, the orresponing ftor for the other roots will be qurti expression. The seon step is then solving qurti eqution. Reltions between roots n oeffiients of ubi eqution Assuming the roots of x + bx + x + = to be α, β n γ We n re-write the eqution s x ( α)( x β)( x γ) = Thus x + bx + x + ( x α)( x β)( x γ) We see tht b α + β + γ = αβ + βγ + γα = αβγ = If one non-zero root γ is foun, b We will hve α + β = γ n αβ = γ With these, we n solve the remining qurti eqution, perhps, s effetive s by inspetion If one root γ is foun to be zero, then =, the eqution is tully x + bx + x = whih is xx ( + bx+ ) = The other root is given by the qurti eqution x bx + + =

3 Exerise (without solution tthe) 1. Given the qurti eqution x + bx + = Show tht if n hve ifferent signs, the eqution will hve rel roots, but the onverse of this sttement is not true. Given the qurti eqution x + bx + = Stte the onitions for it to hve (i) two roots of ifferent signs (ii) two roots of the sme sign. Given two equtions u+ v= h, uv= k. Show how to solve for u n v. Show tht when L is lrge vlue, g( L) = L + bl + L+ hs the sme sign s L, g( L) n g( L) hve ifferent signs 5. Show tht x = 1 is root of the eqution f x = x x + x+ = ( ) 8 Show tht the sum of the other two roots is 1 n tht their prout is Hene write own the qurti eqution for the other two roots 6. Given tht p q= 7 n pq= 1. Solve for pq, by using the ie of sum of roots n prout of roots. (two sets of nswers)

4 Exerise (with solution tthe) 1 Given the qurti eqution x + bx + = Show tht if n hve ifferent signs, the eqution will hve rel roots, but the onverse of this sttement is not true Consier the isriminnt = b. If is negtive, then = b is positive. On the other hn, = b is positive oes not require tht is negtive Given the qurti eqution x + bx + = Stte the onitions for it to hve (i) two roots of ifferent signs (ii) two roots of the sme sign (i) Assume α, β to be the two roots, then αβ = < if n hve ifferent signs, this in turn ensures rel roots for the eqution So the onition is tht n hve ifferent signs. (ii) Assume α, β to be the two roots, then αβ = > Further, to ensure rel roots, = b So the onitions re tht n hve sme sign, n tht = b Given two equtions u+ v= h, uv= k. Show how to solve for u n v (I) We my view uv, s two roots, n so we hve Sum of roots = h, prout of roots= k, We then solve x hx+ k = k k (II) By substitution v =, u+ = h, so u hu+ k = u u. Show tht when L is lrge vlue, g( L) = L + bl + L+ hs the sme sign s L, g( L) n g( L) hve ifferent signs b gl ( ) = L( + ) L + L + L, when L is lrge, b + + is very smll s L L L b ompre to, so the sign of is the sign of, L L L the sign of g( L ) is the sign of L Similrly, the sign of g( L) is the sign of ( L) = L So g( L) n g( L) hve ifferent sign, when L is lrge 5. Show tht x = 1 is root of the eqution f x = x x + x+ = ( ) 8

5 Show tht the sum of the other two roots is 1 n tht their prout is Hene write own the qurti eqution for the other two roots We fin tht f (1) =, So, x = 1 is root of the eqution f( x) = x 8x + x+ = Let the three roots be α, β n γ, with γ = 1 Then α + β + γ = An αβγ = Therefore α + β = 1, αβ = The qurti eqution for the remining two roots is x x =, whih is the sme s x x = 6. Given tht p q= 7 n pq= 1. Solve for p, q by using the ie of sum of roots n prout of roots. (two sets of nswers) Let α = p n β = q, then we hve α + β = 7 n αβ = 1 The qurti eqution for α, β is x 7x+ 1= We get ( α, β ) = (, 5) or (5, ), so ( p, q) = (, -5) or (5, -) 5