Question

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

Answer 1

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

Note: We can also find the roots of a quadratic equation by using the formula, which is,
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Where, the quadratic equation is \[a{x^2} + bx + c = 0.\] 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.