Question

(iii) Solve the quadratic equation x(x - 1) = 1.​

Answer 1

Required Answer:-

Given:

• x(x - 1) = 1

To find:

• The value of x.

Solution:

Given,

$\rm x(x - 1) = 1$

$\rm \implies {x}^{2} - x = 1$

$\rm \implies {x}^{2} - x - 1 = 0$

Now, it is in the standard form.

Here,

$\rm a = 1$

$\rm b = - 1$

$\rm c = - 1$

Therefore,

$\rm x_{1,2} = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$

$\rm = \dfrac{1 \pm \sqrt{ {( - 1)}^{2} - 4 \times1 \times ( - 1)}}{2 \times 1}$

$\rm = \dfrac{1 \pm \sqrt{1 + 4}}{2 \times 1}$

$\rm = \dfrac{1 \pm \sqrt{5}}{2}$

Hence,

$\rm x_{1} = \dfrac{1 + \sqrt{5} }{2}$

$\rm x_{2} = \dfrac{1 - \sqrt{5} }{2}$

So, the roots of the equation are (1 - √5)/2 and (1 + √5)/2

Answer:

• The roots of the equation are (1 + √5)/2 and (1 - √5)/2

Answer 2

$\sf{x\left(x-1\right)=1}$

$\implies\sf{x^2-x-1=1-1}$

$\implies\sf{x^2-x-1=0}$

$\bf{Quadratic\:Equation\:Formula:}$

$\tt{For\:a\:quadratic\:equation\:of\:the\:form\:}\:ax^2+bx+c=0\tt{\:the\:solutions\:are\:}$

$\tt{x_{1,\:2}=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}}$

In our case, we have :

$\sf{a=1,\:b=-1,\:c=-1}$

$\sf{\displaystyle x_{1,\:2}=\frac{-\left(-1\right)\pm \sqrt{\left(-1\right)^2-4\cdot \:1\cdot \left(-1\right)}}{2\cdot \:1}}$

$\implies\sf{\displaystyle x_{1,\:2}=\frac{-\left(-1\right)\pm \sqrt{5}}{2\cdot \:1}}$

$\implies\sf{\displaystyle x_1=\frac{-\left(-1\right)+\sqrt{5}}{2\cdot \:1},\:x_2=\frac{-\left(-1\right)-\sqrt{5}}{2\cdot \:1}}$

$\implies\bf{\displaystyle x=\frac{1+\sqrt{5}}{2},\:x=\frac{1-\sqrt{5}}{2}}$