The set of values of $x$ which satisfy the inequality $0.7^{(2x^2 - 3x + 4)} < 0.343$ is

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Concept:

$(0.7)^a < (0.7)^3$

Since $0.7 < 1$, the inequality flips when comparing exponents, so $a > 3$

Why? A base less than $1$ raised to a higher power gives a smaller result. For example, $(0.7)^4 = 0.2401 < 0.343 = (0.7)^3$. So for $(0.7)^a$ to be less than $(0.7)^3$, we need $a > 3$.

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Since $0.343 = (0.7)^3$, the inequality becomes:

$(0.7)^{2x^2 - 3x + 4} < (0.7)^3$

Flipping the inequality (base $< 1$):

$2x^2 - 3x + 4 > 3$

$2x^2 - 3x + 1 > 0$

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Factorizing:

$2x^2 - 2x - x + 1 > 0$

$2x(x - 1) - 1(x - 1) > 0$

$(2x - 1)(x - 1) > 0$

Roots are $x = \dfrac{1}{2}$ and $x = 1$

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We need the product to be positive, so we check the signs across the number line:

Region	Sign of $(2x-1)$	Sign of $(x-1)$	Product
$x < \dfrac{1}{2}$	$-$	$-$	$+$ ✅
$\dfrac{1}{2} < x < 1$	$+$	$-$	$-$ ❌
$x > 1$	$+$	$+$	$+$ ✅

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$\boxed{x \in \left(-\infty,\ \dfrac{1}{2}\right) \cup \left(1,\ \infty\right)}$