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For what values of j does the equation (2x + 7)(x + 43) = -5 + jx have exactly one real solution? Express your answer as a list of numbers, separated by commas.

Jan 31, 2021

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For what values of j does the equation (2x + 7)(x + 43) = -5 + jx have exactly one real solution? Express your answer as a list of numbers, separated by commas.

Hello Guest!

$$(2x + 7)(x + 43) = -5 + jx\\ 2x^2+86x+7x+301+5-jx=0\\ 2x^2+(93-j)x+306=0$$

$$x = \large {93+j \pm \sqrt{ (93+j)^2-4\cdot 2\cdot 306} \over 2\cdot 306}$$

$$(93+j)^2-4\cdot 2\cdot 306\geq 0\\ 8649+186j+j^2-2448\geq 0\\ j^2+186j+6101\geq 0$$

$$j=-93\pm \sqrt{8649-6101}\\ j=-93\pm 50.478$$

$$j\in \{-42.522,-149.477\}$$

If the value of $$j\in \{-42.522,-149.477\}$$,

the equation (2x + 7) (x + 43) = -5 + jx has exactly one real solution.

If the value of

$$j\in \mathbb R\ |-149.477\leq j\leq -42.522$$,

the equation (2x + 7) (x + 43) = -5 + jx has  two real solutions each.

!

Jan 31, 2021
edited by asinus  Jan 31, 2021
edited by asinus  Jan 31, 2021
edited by asinus  Jan 31, 2021
edited by asinus  Jan 31, 2021