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при каких значениях $a$ неравенство $ax^2+4x>1-3a$ выполняется при всех значениях $х$ ?

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Factor polynomials

Full solution

Q. при каких значениях

a

неравенство

ax^2+4x>1-3a

выполняется при всех значениях

х

?

Find the discriminant: Step Title: Find the discriminant $\newline$ Calculation: The discriminant of a quadratic inequality $ax^2 + bx + c > 0$ is $b^2 - 4ac$ . Here, $a = a$ , $b = 4$ , $c = 3a - 1$ . So, discriminant = $4^2 - 4(a)(3a - 1)$ . $\newline$ Step Output: Discriminant = $16 - 4a(3a - 1)$
Set discriminant less than $0$ : Step Title: Set the discriminant less than $0$ $\newline$ Calculation: For the inequality to hold for all $x$ , the discriminant must be less than $0$ . So, $16 - 4a(3a - 1) < 0$ . $\newline$ Step Output: $16 - 12a^2 + 4a < 0$
Solve the inequality: Step Title: Solve the inequality $\newline$ Calculation: $12a^2 - 4a - 16 > 0$ . Divide by $4$ to simplify: $3a^2 - a - 4 > 0$ . $\newline$ Step Output: $3a^2 - a - 4 > 0$
Factor the quadratic inequality: Step Title: Factor the quadratic inequality $\newline$ Calculation: To factor $3a^2 - a - 4$ , we need two numbers that multiply to $-12$ $(3 \times -4)$ and add to $-1$ . The numbers are $-4$ and $3$ . So, $(3a + 4)(a - 1) > 0$ . $\newline$ Step Output: $(3a + 4)(a - 1) > 0$

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