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Question: If a and b are the roots of the quadratic equation \({{x}^{2}}+px+12=0\) with the condition \(a-b=1\...

If a and b are the roots of the quadratic equation x2+px+12=0{{x}^{2}}+px+12=0 with the condition ab=1a-b=1, then the value of ‘p’ is:
(). 1
(). 7
(). -7
(). +7 or -7

Explanation

Solution

Hint: We will use the formula of sum of roots and product of roots. And with the help of that we can find the value of a + b and ab. Then we will use the formula (ab)2=(a+b)24ab{{\left( a-b \right)}^{2}}={{\left( a+b \right)}^{2}}-4ab and then we will substitute all the values that we know to find the value of p.

Complete step-by-step answer:

If the equation is ax2+bx+c=0a{{x}^{2}}+bx+c=0 and roots are α\alpha and β\beta ,
Then the formula for sum of roots is: α+β=ba\alpha +\beta =\dfrac{-b}{a}
The formula for product of roots is: αβ=ca\alpha \beta =\dfrac{c}{a}
Using the above formula for the equation x2+px+12=0{{x}^{2}}+px+12=0 we get,
The sum of roots as: a+b=p........(1)a+b=-p........(1)
The product of roots as: ab=12.........(2)ab=12.........(2)
Now we have been given that a – b = 1
Now using the formula (ab)2=(a+b)24ab{{\left( a-b \right)}^{2}}={{\left( a+b \right)}^{2}}-4ab and substituting the values from (1) and (2) we get,
(1)2=(p)24(12) 1=p248 p2=49 p=±7 \begin{aligned} & {{\left( 1 \right)}^{2}}={{\left( -p \right)}^{2}}-4\left( 12 \right) \\\ & 1={{p}^{2}}-48 \\\ & {{p}^{2}}=49 \\\ & p=\pm 7 \\\ \end{aligned}
Hence, from this we get the value of p as +7 and -7.
Hence, the correct answer is option (d).

Note: The formula for sum of roots and product of roots must be kept in mind. And how we have used (ab)2=(a+b)24ab{{\left( a-b \right)}^{2}}={{\left( a+b \right)}^{2}}-4ab this formula to solve this question easily. One can also solve this question by finding the value of a and b in terms of p and then substituting it in a – b = 1, and from there also we can find the value of p.