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Question: The common root of \[{x^2} + 5x + 6 = 0{\text { and }} {x^2} - 8x + 15 = 0\] is the root of \[{x^2} ...

The common root of x2+5x+6=0 and x2−8x+15=0{x^2} + 5x + 6 = 0{\text { and }} {x^2} - 8x + 15 = 0 is the root of x2+4x+q=0{x^2} + 4x + q = 0 then value of q is
A. 21
B. 41
C. -21
D. -41

Explanation

Solution

Hint: At first, find the roots of quadratic equations individually by using the factorisation method. Then, we will find the common root which is also said to be the root of the equation x2+4x+q=0{x^2} + 4x + q = 0. So, to find the value 'q' put the value of 'x' as the common root to get the answer.

Complete step-by-step answer:
In the question, we have been given two quadratic equations x2+5x+6=0 and x2−8x+15=0{x^2} + 5x + 6 = 0{\text { and }} {x^2} - 8x + 15 = 0. Further, we are said that, the common root of the two quadratic equations is also a root of equation x2+4x+q=0.{x^2} + 4x + q = 0.
We will first find the roots of the respective quadratic equations x2−5x+6=0 and x2−8x+15=0{x^2} - 5x + 6 = 0{\text { and }}{x^2} - 8x + 15 = 0. We will be using the factorisation method to get the roots as explained below.
So, for the equation x2−5x+6=0{x^2} - 5x + 6 = 0 we can rewrite it as x2−2x−3x+6=0{x^2} - 2x - 3x + 6 = 0 which can be also written as x(x−2)−3(x−2)=0x\left( {x - 2} \right) - 3\left( {x - 2} \right) = 0 and hence factorized as(x−3)(x−2)=0\left( {x - 3} \right)\left( {x - 2} \right) = 0. Thus, the roots of equations are 2 and 3.
Now, we will find for the equation x2−8x+15=0{x^2} - 8x + 15 = 0 which we can rewrite it as x2−3x−5x+15=0{x^2} - 3x - 5x + 15 = 0 which can also be written as x(x−3)−5(x−3)=0x\left( {x - 3} \right) - 5\left( {x - 3} \right) = 0and hence can be factorized as (x−5)(x−3)=0.\left( {x - 5} \right)\left( {x - 3} \right) = 0. Thus, the roots of equation are 5 and 3.
The roots of equation x2−5x+6=0{x^2} - 5x + 6 = 0 is 2 and 3 and roots of equation x2−8x+15=0{x^2} - 8x + 15 = 0 is 5 and 3, thus, by observing we can say that, the common root between them is 3.
We were said that, the common root of the equations x2−5x+6=0 and x2−8x+15=0{x^2} - 5x + 6 = 0{\text { and }} {x^2} - 8x + 15 = 0 is also a root of x2+4x+q=0.{x^2} + 4x + q = 0.
The common root of x2−5x+6=0 and x2−8x+15=0{x^2} - 5x + 6 = 0{\text { and }} {x^2} - 8x + 15 = 0 is 3, thus, we can say 3 is a root of x2+4x+q=0{x^2} + 4x + q = 0 or we can say that 3 satisfies the quadratic equation x2+4x+q=0.{x^2} + 4x + q = 0.
So, now to find the value of 'q' we will substitute x as 3, so we get,
(3)2+4×3+q=0 ⇒9+12+q=0\begin{array}{l}{\left( 3 \right)^2} + 4 \times 3 + q = 0\\\ \Rightarrow 9 + 12 + q = 0\end{array}
Now on simplifying, we get,
21+q=0⇒q=−2121 + q = 0 \Rightarrow q = - 21
Hence, the correct option is C.

Note: We can also find the roots of a quadratic equation by using the formula, which is,
x=−b±b2−4ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}
Where, the quadratic equation is ax2+bx+c=0.a{x^2} + bx + c = 0. But this might be confusing and some students might get the formula wrong, so for simple quadratic equations like the ones in this question, the factorisation method is better.