The roots of $x^2 + px + 12 = 0$ differ by 1, and $p$ is positive. What is the value of $p$ ?

7Correct answerYour answer

\sqrt{13}

Correct answerYour answer

-7

Correct answerYour answer49Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer A

Correct answer: A

For roots $\alpha, \beta$ : $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = p^2 - 48$ .
The difference is 1, so $p^2 - 48 = 1$ and $p^2 = 49$ .
$p = \pm 7$ ; since $p$ is positive, $p = 7$ . (The roots are $-3$ and $-4$ .)

Worth noting

$-7$ also gives roots differing by 1 but is ruled out by the condition. $\sqrt{13}$ comes from using $\alpha\beta$ instead of $4\alpha\beta$ , and 49 is $p^2$ .

Concept: Difference of roots from sum and product
Typical timeTypical: 120 s
Your timeYours: —

The rule this testsRoots and their nature

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Ten questions from Linear and Quadratic Equations, at your own pace.

The roots of x2+px+12=0x^2 + px + 12 = 0x2+px+12=0 differ by 1, and ppp is positive. What is the value of ppp?

Explanation

The roots of $x^2 + px + 12 = 0$ differ by 1, and $p$ is positive. What is the value of $p$ ?