Question

what is the solution to the equation x^2 - 5x = -4

Answer 1

Answer:

$x=1,\ 4$

Step-by-step explanation:

Method 1: Solving by factorization

$x^2-5x=-4\\\text{or, }x^2-5x+4=0\\\text{or, }x^2-4x-x+4=0\\\text{or, }x(x-4)-1(x-4)=0\\\text{or, }(x-4)(x-1)=0\\\text{i.e. }x=4\text{ or }1$

Method 2: Solving by the quadratic formula

$x^2-5x=-4\\\text{or, }x^2-5x+4=0\\\text{Comparing with }ax^2+bx+c=0,\ a=1,\ b=-5\text{ and }c=4.\\\text{Using the quadratic formula,}\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-5)\pm\sqrt{(-5)^2-4(1)(4)}}{2(1)}=\dfrac{5\pm\sqrt{25-16}}{2}\\\\\text{or, }x=\dfrac{5\pm3}{2}$

$\text{Taking positive sign,}\\x=\dfrac{5+3}{2}=4\\\\\\\text{Taking negative sign,}\\\\x=\dfrac{5-3}{2}=1\\\\\therefore\ x=4,\ 1$

Method 3: Completing square

$x^2-5x=-4\\\text{or, }4(x^2-5x)=4(-4)\\\text{or, }4x^2-20x=-16\\\text{or, }(2x)^2-2(2x)(5)+5^2=5^2-16=25-16\\\text{or, }(2x-5)^2 =9\\\text{or, }(2x-5)^2=3^2\\\text{or, }2x-5=\pm3\\\text{or, }2x=\pm3+5\\\\\text{Taking positive sign,}\\2x=3+5=8\\\text{or, }x=4\\\\\text{Taking negative sign,}\\2x=-3+5=2\\\text{or, }x=1$