The diagram shows a curve with equation $$\(y=4 x^{\frac{1}{2}}-2 x\)$$ for $$\(x \geqslant 0\)$$, and a straight line with equation $$\(y=3-x\)$$. The curve crosses the $$\(x\)$$-axis at $$\(A(4,0)\)$$ and crosses the straight line at $$\(B\)$$ and $$\(C\)$$. Find, by calculation, the $$\(x\)$$-coordinates of $$\(B\)$$ and $$\(C\)$$. .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................
Exam No：9709_w20_qp_11 Year：2020 Question No：12(a)
Answer:
$\begin{array}{l}4 x^{\frac{1}{2}}-2 x=3-x \rightarrow x-4 x^{\frac{1}{2}}+3(=0) \\ \left(x^{\frac{1}{2}}-1\right)\left(x^{\frac{1}{2}}-3\right)(=0) \text { or }(u-1)(u-3)(=0) \\ x^{\frac{1}{2}}=1,3 \\ x=1,9\end{array}$
Alternative method for question 12(a)
$\begin{array}{l}\left(4 x^{\frac{1}{2}}\right)^{2}=(3+x)^{2} \\ 16 x=9+6 x+x^{2} \rightarrow x^{2}-10 x+9(=0) \\ (x-1)(x-9)(=0) \\ x=1,9\end{array}$
Alternative method for question 12(a)
$\begin{array}{l}\left(4 x^{\frac{1}{2}}\right)^{2}=(3+x)^{2} \\ 16 x=9+6 x+x^{2} \rightarrow x^{2}-10 x+9(=0) \\ (x-1)(x-9)(=0) \\ x=1,9\end{array}$
Knowledge points:
1.1.4 solve by substitution a pair of simultaneous equations of which one is linear and one is quadratic
1.1.5 recognise and solve equations in which are quadratic in some function of
Solution:
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