Question

38. Solve : 8x°- 2x² - 7x+3 = 0​

Answer 1

Answer:

Before solving this question please remember that (x)^0 = 1 always.

In fact, any number raised to the power zero is equal to 1.

So, the given equation will become

$1 - 2 {x}^{2} - 7x + 3 = 0 \\ - 2 {x }^{2} - 7x + 3 + 1 = 0 \\ - 2 {x}^{2} - 7x + 4 = 0 \\ transposing \: all \: the \: terms \: from \: one \: side \: to \: another \: the \: signs \: of \: the \: terms \: will \: change \: \\ 0 = 2 {x}^{2} + 7x - 4 \\ this \: can \: now \: \: written \: as \\ 2 {x }^{2} + 7x - 4 = 0 \\ 2 {x}^{2} + 8x - x - 4 = 0 \\ 2 \times x \times x + 2 \times x \times 4 - 1 \times x - 1 \times 4 = 0 \\ 2x \times (x + 4) - 1 \times (x + 4) = 0 \\ (x + 4)(2x - 1) = 0$