Quadratic and simultaneous equations — Paper 1 long questions

(a) Equal roots means $\Delta=0$: $b^{2}-4(4)(49)=0$, so $b^{2}=784$ and $b=\pm 28$. As $b<0$, $b=-28$.

(b) Then $p(x)=4x^{2}-28x+49=(2x-7)^{2}$, so $(2x-7)^{2}=36$ gives $2x-7=\pm 6$. The $x$-intercepts are $x=\tfrac{13}{2}$ and $x=\tfrac{1}{2}$. Spotting the perfect square is faster here than the quadratic formula.

(a) $\alpha+\beta=-\tfrac{c}{2}$ and $\alpha\beta=-3$, so $\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2\alpha\beta=\dfrac{c^{2}}{4}+6$.

(b) $\dfrac{c^{2}}{4}+6=22$ gives $c=\pm 8$. The new roots have sum $(\alpha+\beta)+2$ and product $\alpha\beta+(\alpha+\beta)+1$. For $c=8$: sum $-2$, product $-6$, giving $x^{2}+2x-6=0$. For $c=-8$: sum $6$, product $2$, giving $x^{2}-6x+2=0$. Neither case needs the roots themselves — they are irrational, and finding them wastes time.

(a) Eliminating $y$ gives $x^{2}+(2k-2)x+(k-3)=0$. Its discriminant is $(2k-2)^{2}-4(k-3)=4k^{2}-12k+16=4\left[\left(k-\tfrac{3}{2}\right)^{2}+\tfrac{7}{4}\right]>0$ for every real $k$, so there are always two distinct intersection points.

(b) $a+b=2-2k$ and $ab=k-3$, so $(a-b)^{2}=(a+b)^{2}-4ab=(2-2k)^{2}-4(k-3)=4k^{2}-12k+16$.

(c) No. Both points lie on $L$, whose slope is $1$, so $AB=\sqrt{2}\,|a-b|$. Then $AB=2\sqrt{2}$ forces $(a-b)^{2}=4$, i.e. $4k^{2}-12k+16=4$, i.e. $k^{2}-3k+3=0$. That equation has discriminant $9-12=-3<0$, so no real $k$ gives $AB=2\sqrt{2}$. The slope factor is where most marks are lost.

(a) Let $L:\;y=mx$. Eliminating $y$ gives $x^{2}-(2+m)x+5=0$. The mid-point of $PQ$ has $x$-coordinate $\dfrac{2+m}{2}=3$, so $m=4$ and $L:\;y=4x$. (Check: $\Delta=36-20=16>0$, so the two points are indeed distinct.)

(b) $x^{2}-6x+5=0$ gives $x=1$ or $x=5$, so $P=(1,4)$ and $Q=(5,20)$. The sum of roots is the whole of part (a) — solving the quadratic first is unnecessary work.

(a) $\Delta=4k^{2}-4(k+6)>0$, so $k^{2}-k-6>0$, i.e. $(k-3)(k+2)>0$. Hence $k<-2$ or $k>3$.

(b) Non-real roots need $\Delta<0$, so $-2<k<3$. Sum and product still apply: $\alpha^{2}+\beta^{2}=(2k)^{2}-2(k+6)=4k^{2}-2k-12$. This is least at $k=\tfrac{1}{4}$, which lies in the range, giving $-\tfrac{49}{4}$. A negative value is not an error — $\alpha$ and $\beta$ are not real, so $\alpha^{2}+\beta^{2}$ need not be positive.

(a) Let $t=x^{2}+2x$. Then $t^{2}-t-6=0$, so $(t-3)(t+2)=0$ and $t=3$ or $t=-2$. From $x^{2}+2x-3=0$: $x=1$ or $x=-3$. From $x^{2}+2x+2=0$: $\Delta=4-8=-4<0$, no real root. So the real roots are $x=1$ and $x=-3$.

(b) Disagree. The equation is quartic, so it has four roots in total, but two of them are non-real — they come from $x^{2}+2x+2=0$, whose discriminant is negative. Only two roots are real. Checking the discriminant of each branch is the step that settles it.

(a) Let the sides be $x$ cm and $y$ cm. Then $x+y=22$ and $xy=105$, so $x$ and $y$ are the roots of $t^{2}-22t+105=0$, i.e. $(t-7)(t-15)=0$. The length is $15$ cm and the width is $7$ cm.

(b) Not possible. The sides would satisfy $t^{2}-22t+125=0$, whose discriminant is $22^{2}-4(125)=-16<0$. There are no real side lengths, so no such rectangle exists. (The largest area a $44$ cm perimeter allows is $121\ \text{cm}^{2}$, from an $11\times 11$ square.)

(a) Sum of roots: $q+3q=-a$, so $q=-\tfrac{a}{4}$. Product of roots: $q(3q)=3q^{2}=b$, so $3\left(\tfrac{a^{2}}{16}\right)=b$, i.e. $3a^{2}=16b$.

(b) Substituting $y=mx$ into $C$ gives $(1+m^{2})x^{2}+(16m-12)x+15=0$. Since $P$ and $R$ lie on a line through $O$, $OP:OR=1:3$ means their $x$-coordinates are in the ratio $1:3$, so the roots are $q$ and $3q$. Applying (a) in the form $3B^{2}=16AC$: $3(16m-12)^{2}=16(1+m^{2})(15)$, which simplifies to $11m^{2}-24m+4=0$. Hence $m=2$ or $m=\tfrac{2}{11}$. Both are valid — the ratio condition does not say which point is nearer $O$.