1fully worked

Solve $x^2+3x+1=0$ in surd form.

1$a=1,\ b=3,\ c=1$Write $a$, $b$, $c$

2$x=\dfrac{-3\pm\sqrt{9-4}}{2}$Substitute into $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

3$=\dfrac{-3\pm\sqrt{5}}{2}$Simplify under the root

Check - Why is the first term $-3$ and not $+3$?

A the formula starts with $-b$, and $b=3$B because $3$ is oddC a typing error – it should be $+3$D because $c=1$

2you finish it

Solve $x^2+5x+3=0$ in surd form.

1$a=1,\ b=5,\ c=3$Write $a$, $b$, $c$

2$x=\dfrac{-5\pm\sqrt{25-12}}{2}$Substitute into $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

3Simplify under the root

Check - What is $b^2-4ac$ here?

A $25+12=37$B $25-12=13$C $5-12=-7$D $10-12=-2$

3you finish it

Solve $x^2-x-1=0$ in surd form.

1$a=1,\ b=-1,\ c=-1$Write $a$, $b$, $c$

2Substitute into $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

3Simplify under the root

Check - Why does $-4ac$ become $+4$?

A $-4\times1\times(-1)=+4$ (two negatives)B a mistake – it should be $-4$C because $b=-1$D because the root is positive

4you do it all

Solve $x^2+3x-1=0$ in surd form.

1Write $a$, $b$, $c$

2Substitute into $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

3Simplify under the root

Check - The two roots come from the $\pm$. What does that mean here?

A only take the $+$ signB there are two roots: $\dfrac{-3+\sqrt{13}}{2}$ and $\dfrac{-3-\sqrt{13}}{2}$C the answer is a single numberD the $\pm$ can be ignored