Question

Solve the following equation and write your answer correct to two significant figures:
${{\left( x-1 \right)}^{2}}-3x+4=0$

Answer 1

- Hint: Put x- 1 = t and hence form a quadratic equation in t. Use the method of completing the square to get the roots of the equation. Alternatively, use the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, where a is the coefficient of ${{x}^{2}}$, b the coefficient of x and c the constant term in the equation to get the roots of the equation.

Complete step-by-step solution -
Put x-1 = t
Hence, we have x = t+1
Now, we have
${{\left( x-1 \right)}^{2}}-3x+4={{\left( t+1-1 \right)}^{2}}-3\left( t+1 \right)+4={{t}^{2}}-3t+1$
Hence, the equation becomes
${{t}^{2}}-3t+1=0$
We solve the quadratic equation using the method of completing the square.
Step 1: Make the coefficient of ${{t}^{2}}$ as 1.
The given equation already has the coefficient of ${{t}^{2}}$ as 1.
Step 2: Write the linear term in $2at$ form
We have ${{t}^{2}}-3t+1={{t}^{2}}-2\times \dfrac{3}{2}t+1=0$, which is in the required form
Step 3: Add ${{a}^{2}}$ on both sides of the equation
${{t}^{2}}-2\times \dfrac{3}{2}t+{{\left( \dfrac{3}{2} \right)}^{2}}+1={{\left( \dfrac{3}{2} \right)}^{2}}$
Step 4: Use one of the identities ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ and ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$and hence solve the equation.
We have ${{t}^{2}}-2\times \dfrac{3}{2}t+{{\left( \dfrac{3}{2} \right)}^{2}}={{\left( t-\dfrac{3}{2} \right)}^{2}}$
Hence the equation becomes
${{\left( t-\dfrac{3}{2} \right)}^{2}}+1=\dfrac{9}{4}$
Subtracting 1 from both sides, we get
${{\left( t-\dfrac{3}{2} \right)}^{2}}=\dfrac{5}{4}$
Hence, we have
$t-\dfrac{3}{2}=\pm \dfrac{\sqrt{5}}{2}$
Taking the positive sign, we get
$t-\dfrac{3}{2}=\dfrac{\sqrt{5}}{2}\Rightarrow t=\dfrac{3+\sqrt{5}}{2}$
Taking the negative sign, we get
$t-\dfrac{3}{2}=\dfrac{-\sqrt{5}}{2}=\dfrac{3-\sqrt{5}}{2}$
Reverting to the original variable, we have
$x=t+1=\dfrac{5+\sqrt{5}}{2}\text{ or }\dfrac{5-\sqrt{5}}{2}$
Hence, we have
$x=3.62\text{ or }1.38$, which is the required solution to correct up to two significant places.

Note: Verification:
If the sum of roots = $\dfrac{-b}{a}$ and the product of roots $=\dfrac{c}{a}$, then the given roots are the roots of the equation (a,b,c) have their usual meanings.
We have
${{\left( x-1 \right)}^{2}}-3x+4=0\Rightarrow {{x}^{2}}-5x+5=0$
Hence a = 1, b =-5 and c =5
Sum of roots $=\dfrac{5+\sqrt{5}}{2}+\dfrac{5-\sqrt{5}}{2}=5=\dfrac{-\left( -5 \right)}{1}=-\dfrac{b}{a}$
Product of roots $=\dfrac{\left( 5+\sqrt{5} \right)}{2}\dfrac{\left( 5-\sqrt{5} \right)}{2}=\dfrac{25-5}{4}=\dfrac{20}{4}=5=\dfrac{c}{a}$
Hence our answer is verified to be correct.