(CIE 0606/2020/w/11/Q3)
(a) Write √p(qr2)13(q3p)−1r3 in the form paqbrc, where a,b and c are constants. [3]
(b) Solve 6x23−5x13+1=0.
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*********math solution*************
(a) p12(qr2)13(q3p)−1r3=p12q13r23(q3p)r−3=p12+1q13+3r23−3=p32q103r73
(b) 6x23−5x13+1=06(x13)2−5(x13)+1=0(3x13−1)(2x13−1)=03x13=1(or)2x13=1x13=13(or)x13=12x=127(or)x=18
**********end math solution********************
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