Question

3.If the difference of roots of the quadratic equation x2+x+12=0 is 1, the positive value of k is A) -7 (b) 7 (c) 4 (d) 8​

Answer 1

option B

There is mistake in question

quadratic equation =

x^2+kx+12=0 is correct

so,

Here

a=1 , b=k and c=12

$\alpha + \beta = \frac{ - b}{a} = \frac{ - k}{1} = - k$

and

$\alpha \beta = \frac{c}{a} = 12$

And also given that

Difference of root is 1 means ,

$\alpha - \beta = 1$

$\alpha - \beta = \sqrt{( \alpha + \beta ) ^{2} - 4 \alpha \beta }$

here put the values and we get the answer

soln refers to the attachment.

Answer 2

p(x)= x^2+kx+12=0

here a=1 b= k and c=12

given that

difference of roots = 1

$\alpha - \beta = 1 \\ = > \sqrt{ {( \alpha + \beta })^{2} - 4 \alpha \beta } = 1$

square both sides then

$({ \alpha + \beta })^{2} - 4 \alpha \beta = 1 \\ = > { - k}^{2} - 4(12) = 1 \\ = > {k}^{2} = 49 \\ = > k = 7$