Q.A quadratic polynomial has one of its zeros $1+\sqrt5$ and it satisfies $p(1)=2$. Find the quadratic polynomial.

For $P(x)=ax^2+bx+c$ ($a\ne0$), completing the square gives the quadratic formula:

Discriminant $D=b^2-4ac$: $D>0$ gives two distinct real roots (parabola crosses the x-axis twice); $D=0$ gives one repeated real root (touches once); $D<0$ gives no real roots (a complex conjugate pair; parabola never meets the x-axis).

Sum and product of roots. For roots $\alpha,\beta$ of $ax^2+bx+c=0$: $\alpha+\beta=-\dfrac ba$, $\alpha\beta=\dfrac ca$. This pair of formulas is the single most useful tool in the chapter for:

• Building an equation from its roots: $x^2-(\alpha+\beta)x+\alpha\beta=0$;
• Transforming roots (reciprocals $1/\alpha,1/\beta$; doubles $2\alpha,2\beta$; negatives $-\alpha,-\beta$) into a new equation without ever solving for $\alpha,\beta$ explicitly;
• Encoding conditions like 'one root is $k$ times the other' or 'the roots differ by a given amount' as algebraic equations in the coefficients, often combined with $(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta$.

Complex roots ($D<0$): $\alpha,\beta=\dfrac{-b\pm i\sqrt{4ac-b^2}}{2a}$, with $i^2=-1$.

Conjugate irrational zeros. If a quadratic has rational coefficients and one zero is $p+q\sqrt d$ ($p,q\in Q$), the other zero must be the conjugate $p-q\sqrt d$.