1 viii) xx==2(x #0 x
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x2– 2x = (–2) (3 – x)
We have:
x2 – 2x = (– 2) (3 – x)
⇒ x2 – 2x = –6 + 2x
⇒ x2 – 2x – 2x + 6 = 0
⇒ x2 – 4x + 6 = 0
Since x2 – 4x + 6 is a quadratic polynomial
∴ x2 – 2x = (–2) (3 – x) is a quadratic equation.
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