Question:
The equations \(ax^{2}+bx+a=0\), \((a, b \in \mathbb{R})\) and \(x^{3}-2x^{2}+2x-1=0\) have two common roots. Then the value of \(a+b\) is ________.
The equations \(ax^{2}+bx+a=0\), \((a, b \in \mathbb{R})\) and \(x^{3}-2x^{2}+2x-1=0\) have two common roots. Then the value of \(a+b\) is ________.
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Factorise the cubic as \((x-1)(x^{2}-x+1)\); real-coefficient quadratics share complex roots in conjugate pairs.
Updated On: Nov 12, 2026
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Solution and Explanation
\(x^{3}-2x^{2}+2x-1=(x-1)(x^{2}-x+1)\). Its roots are \(1\) and the two non-real roots \(\frac{1\pm i\sqrt{3}}{2}\).
The quadratic \(ax^{2}+bx+a=0\) has real coefficients. If it shared the root 1 and one non-real root, the conjugate of that root would also be a root of it, giving three roots for a quadratic, which is impossible. So the two common roots are the non-real pair, and \(ax^{2}+bx+a\) is a multiple of \(x^{2}-x+1\).
Hence \(b=-a\) (the constant term already matches), so \(a+b=0\).
Answer: 0
The quadratic \(ax^{2}+bx+a=0\) has real coefficients. If it shared the root 1 and one non-real root, the conjugate of that root would also be a root of it, giving three roots for a quadratic, which is impossible. So the two common roots are the non-real pair, and \(ax^{2}+bx+a\) is a multiple of \(x^{2}-x+1\).
Hence \(b=-a\) (the constant term already matches), so \(a+b=0\).
Answer: 0
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