Question

find the number of integral values of x so that 2^2x^2-7x+5=1​

Answer 1

2x^2-7x+5=1

2x^2-7x+5-1=0

2x^2-7x+4=0 (2x^2-7x+8)

2x^2-(7+1)x+4

(2x^2-7x)(x+4)

Answer 2

✪ANSWER✪

• The given equation has one integral solution.

★EXPLAINATION★

${2}^{2 {x}^{2} - 7x + 5 } = 1$

${2}^{2 {x}^{2} - 7x + 5} = {2}^{0}$

$2 {x}^{2} - 7x + 5 = 0$

$2 {x}^{2} - 2x - 5x + 5 = 0$

$2x(x - 1) - 5(x - 1) = 0$

$(x - 1)(2x - 5) = 0$

$x = 1 \: or \: \frac{5}{2}$

The only integral solution is x = 1. Hence the equation has one integral solution.