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Exercise 2.4 · Q5

Q.If the difference of the roots of the equation 2x2−(a+1)x+a−1=02x^2-(a+1)x+a-1=0 is equal to their product, then prove that a=2a=2.

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Step 1. From 2x2−(a+1)x+(a−1)=02x^2-(a+1)x+(a-1)=0: α+β=a+12\alpha+\beta=\dfrac{a+1}2, αβ=a−12\alpha\beta=\dfrac{a-1}2.

Step 2. Given condition: α−β=αβ=a−12\alpha-\beta=\alpha\beta=\dfrac{a-1}2.

Step 3. Use (α−β)2=(α+β)2−4αβ(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta: (a−12)2=(a+12)2−4⋅a−12\left(\dfrac{a-1}2\right)^2=\left(\dfrac{a+1}2\right)^2-4\cdot\dfrac{a-1}2. …

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