Consider the function h(x)=x^2-2x+2+ 4x^2-2x+2 , x R . Then the equation h(x)-5=0 has

Substitute t=x^2-2x+2=(x-1)^2+1 . Since (x-1)^2 0 , the new variable satisfies t 1 , and h(x)=t+ 4t. Solve h-5=0 in t . The equation t+ 4t=5 multiplies up to t^2-5t+4=0 , i.e. (t-1)(t-4)=0 , so t=1 or t=4 — both satisfy t 1 , so both are admissible. Count the x -values for each t . Recall t=(x-1)^2+1 . - t=1 : (x-1)^2=0 , giving the single value x=1 . - t=4 : (x-1)^2=3 , giving two values x=1 3 . Altogether there are 1+2=3 solutions. Answer: D.

Quadratics Difficulty 4

TMUA practice: Quadratics, problem 1

A TMUA-calibre quadratics problem at difficulty 4 of 5, with the full worked solution.

The question

Consider the function

h(x)=x^{2}-2x+2+\dfrac{4}{x^{2}-2x+2}

x\in\mathbb{R}

. Then the equation

h(x)-5=0

has

\text{no solution}

\text{only one solution}

\text{exactly two solutions}

\text{exactly three solutions}

The correct answer is highlighted. D

Worked solution

Substitute $t=x^{2}-2x+2=(x-1)^{2}+1$ . Since $(x-1)^{2}\ge0$ , the new variable satisfies $t\ge1$ , and

$h(x)=t+\frac{4}{t}.$

Solve $h-5=0$ in $t$ . The equation $t+\dfrac{4}{t}=5$ multiplies up to $t^{2}-5t+4=0$ , i.e. $(t-1)(t-4)=0$ , so $t=1$ or $t=4$ — both satisfy $t\ge1$ , so both are admissible.

Count the $x$ -values for each $t$ . Recall $t=(x-1)^{2}+1$ .

$t=1$ : $(x-1)^{2}=0$ , giving the single value $x=1$ .
$t=4$ : $(x-1)^{2}=3$ , giving two values $x=1\pm\sqrt{3}$ .

Altogether there are $1+2=3$ solutions.

Answer: D.