Question

${x}^{2} - 5x + 4 =0$
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Answer 1

x^2-5x+4=0.

By the method of splitting the middle term:--

x^2-x-4x+4=0.

x(x-1)-4(x-1)=0.

(x-1)(x-4)=0.

x-1=0 and x-4=0.

x=1 and x=4.

Thus, x=1&4 are the roots is this equation.

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Answer 2

Answer:

THE ROOTS OF THIS EQN ARE----

□x= 4 and 1 ✔✔

Step-by-step explanation:

${x}^{2} - 5x + 4 = 0 \\ first \: we \: use \: middle \: term \: splitting \: \\ {x}^{2} - 4x - x + 4 = 0 \\ x(x - 4) - 1(x - 4) = 0 \\ (x - 1)(x - 4) = 0 \\ x = 1 \: and \: 4 \\ again \: \\ we \: use \: second \: method \: \\ that \: is \: quaratic \: formula \\ d = {b}^{2} - 4ac \\ d = {5}^{2} - 16 \\ d = 25 - 16 \\ d = 9 \\ x = \frac{ - b + \sqrt{d} }{2a} \: \: \: \: \: \: \: \: \: \frac{ - b - \sqrt{d} }{2a} \\ \: \: \: \: = \frac{5 + 3}{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{5 - 3}{2} \\ \: \: \: \: = \frac{8}{2} \: \: \: \: \: \: \: \: \: \: \: \: \frac{2}{2} \\ x = 4 \: and \: 1$