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Exercise 2.4 · Q4

Q.If one root of k(x−1)2=5x−7k(x-1)^2=5x-7 is double the other root, show that k=2k=2 or −25-25.

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Step 1. Expand: k(x−1)2=5x−7⇒kx2−2kx+k=5x−7⇒kx2−(2k+5)x+(k+7)=0k(x-1)^2=5x-7\Rightarrow kx^2-2kx+k=5x-7\Rightarrow kx^2-(2k+5)x+(k+7)=0.

Step 2. Let the roots be rr and 2r2r. Sum: 3r=2k+5k3r=\dfrac{2k+5}{k}, so r=2k+53kr=\dfrac{2k+5}{3k}. Product: 2r2=k+7k2r^2=\dfrac{k+7}{k}, so r2=k+72kr^2=\dfrac{k+7}{2k}.

Step 3. Substitute: (2k+53k)2=k+72k⇒(2k+5)29k2=k+72k\left(\dfrac{2k+5}{3k}\right)^2=\dfrac{k+7}{2k}\Rightarrow\dfrac{(2k+5)^2}{9k^2}=\dfrac{k+7}{2k}.

Step 4. Cross-multiply: 2k(2k+5)2=9k2(k+7)2k(2k+5)^2=9k^2(k+7). Divide by kk (nonzero, else no quadratic): 2(2k+5)2=9k(k+7)2(2k+5)^2=9k(k+7). …

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