
If ${x^2} - hx - 21 = 0$ , ${x^2} - 3hx + 35 = 0$ have a common root then, $h$
A) $ \pm 2$
B) $ \pm 4$
C) $ \pm 6$
D) $ \pm 8$
Answer
219.6k+ views
Hint: A quadratic equation is typically represented as $a{x^2} + bx + c = 0$ , where $a \ne 0$ . Real numbers are denoted by the values $a$ , $b$ and $c$ . If we are given two quadratic equations having a common root, then we derive a formula to find the root that both of the given equations share.
Formula Used: Let $\alpha $ be the common root of the quadratic equations ${a_1}{x^2} + {b_1}x + {c_1} = 0$ and ${a_2}{x^2} + {b_2}x + {c_2} = 0$ . By Crammer’s rule
$\dfrac{{{\alpha ^2}}}{{\left| {\begin{array}{*{20}{c}}
{ - {c_1}}&{{b_1}} \\
{ - {c_2}}&{{b_2}}
\end{array}} \right|}} = \dfrac{\alpha }{{\left| {\begin{array}{*{20}{c}}
{{a_1}}&{ - {c_1}} \\
{{a_2}}&{ - {c_2}}
\end{array}} \right|}} = \dfrac{1}{{\left| {\begin{array}{*{20}{c}}
{{a_1}}&{{b_1}} \\
{{a_2}}&{{b_2}}
\end{array}} \right|}}$ , $\alpha \ne 0$ .
Complete step by step Solution:
We have quadratic equations,
${x^2} - hx - 21 = 0$
${x^2} - 3hx + 35 = 0$
Applying Cramer’s rule to find the common roots of equations (1.1) and (1.2), we get the condition for one root common is
${({c_1}{a_2} - {c_2}{a_1})^2} = ({b_1}{c_2} - {b_2}{c_1})({a_1}{b_2} - {a_2}{b_1})$
We have, ${a_1} = 1$ , ${b_1} = - h$ and ${c_1} = - 21$
${a_2} = 1$ , ${b_2} = - 3h$ and ${c_2} = 35$ .
Substituting, these values in the condition ${({c_1}{a_2} - {c_2}{a_1})^2} = ({b_1}{c_2} - {b_2}{c_1})({a_1}{b_2} - {a_2}{b_1})$
${( - 21 \times 1 - 35 \times 1)^2} = ( - h \times 35 - ( - 3h) \times ( - 21)) \cdot (1 \times ( - 3h) - 1 \times ( - h))$
${( - 21 - 35)^2} = ( - 35h - 63h)( - 3h + h)$
On solving further,
${( - 56)^2} = ( - 98h)( - 2h)$
${h^2} = \dfrac{{3136}}{{196}}$
So, we have $h = \pm 4$
Therefore, the correct option is (B).
Note: Make the coefficient of the 2nd-degree term in the two equations equal and subtract to determine the common root of the two equations. The acquired value of $x$ is the necessary common root. Since imaginary and rational roots always occur in pairs, two different quadratic equations with rational coefficients cannot share a single common root that is complex or irrational.
Formula Used: Let $\alpha $ be the common root of the quadratic equations ${a_1}{x^2} + {b_1}x + {c_1} = 0$ and ${a_2}{x^2} + {b_2}x + {c_2} = 0$ . By Crammer’s rule
$\dfrac{{{\alpha ^2}}}{{\left| {\begin{array}{*{20}{c}}
{ - {c_1}}&{{b_1}} \\
{ - {c_2}}&{{b_2}}
\end{array}} \right|}} = \dfrac{\alpha }{{\left| {\begin{array}{*{20}{c}}
{{a_1}}&{ - {c_1}} \\
{{a_2}}&{ - {c_2}}
\end{array}} \right|}} = \dfrac{1}{{\left| {\begin{array}{*{20}{c}}
{{a_1}}&{{b_1}} \\
{{a_2}}&{{b_2}}
\end{array}} \right|}}$ , $\alpha \ne 0$ .
Complete step by step Solution:
We have quadratic equations,
${x^2} - hx - 21 = 0$
${x^2} - 3hx + 35 = 0$
Applying Cramer’s rule to find the common roots of equations (1.1) and (1.2), we get the condition for one root common is
${({c_1}{a_2} - {c_2}{a_1})^2} = ({b_1}{c_2} - {b_2}{c_1})({a_1}{b_2} - {a_2}{b_1})$
We have, ${a_1} = 1$ , ${b_1} = - h$ and ${c_1} = - 21$
${a_2} = 1$ , ${b_2} = - 3h$ and ${c_2} = 35$ .
Substituting, these values in the condition ${({c_1}{a_2} - {c_2}{a_1})^2} = ({b_1}{c_2} - {b_2}{c_1})({a_1}{b_2} - {a_2}{b_1})$
${( - 21 \times 1 - 35 \times 1)^2} = ( - h \times 35 - ( - 3h) \times ( - 21)) \cdot (1 \times ( - 3h) - 1 \times ( - h))$
${( - 21 - 35)^2} = ( - 35h - 63h)( - 3h + h)$
On solving further,
${( - 56)^2} = ( - 98h)( - 2h)$
${h^2} = \dfrac{{3136}}{{196}}$
So, we have $h = \pm 4$
Therefore, the correct option is (B).
Note: Make the coefficient of the 2nd-degree term in the two equations equal and subtract to determine the common root of the two equations. The acquired value of $x$ is the necessary common root. Since imaginary and rational roots always occur in pairs, two different quadratic equations with rational coefficients cannot share a single common root that is complex or irrational.
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