Question

25. Find the value(s) of k so that the equations x² - 11x + k = 0 and x² - 14x + 2k = 0 may have
a common root.
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Answer 1

Answer:

k = 75/4

Step-by-step explanation:

For roots to be equal discriminant must be 0.

i.e.

${b}^2 - 4ac = 0$

comparing equations x² - 11x + k = 0 and x² - 14x + 2k = 0 with ax² + b + c = 0

in x² - 11x + k = 0

a = 1, b = -11, c = k

and in x² - 14x + 2k = 0

a = 1, b = -14, c = 2k

for x² - 11x + k

${b}^{2} - 4ac = 0 \\ { - 11}^{2} - 4(1)(k) = 0 \\ 121 - 4k = 0$

for x² - 14x + 2k

${b}^{2} - 4ac =0 \\ { - 14}^{2} - 4(1)(2k) = 0 \\ 196 - 8k = 0$

we can write,

121 - 4k = 196 - 8k

8k - 4k = 196 - 121

4k = 75

k = 75/4

Answer 2

$\large\underline{\sf{Solution-}}$

Given quadratic equations are

$\rm :\longmapsto\: {x}^{2} - 11x + k = 0 - - (1)$

and

$\rm :\longmapsto\: {x}^{2} - 14x + 2k = 0 - - - (2)$

Let assume that 'y' be the common root.

Thus,

y must satisfy equation (1) and equation (2),

$\rm :\longmapsto\: {y}^{2} - 11y + k = 0 - - (3)$

and

$\rm :\longmapsto\: {y}^{2} - 14y + 2k = 0 - - - (4)$

Now,

Subtracting equation (4) from equation (3), we get

$3y - k = 0$

$\bf\implies \: y= \dfrac{k}{3} - - - (5)$

Now

Substituting the value of y in equation (3), we get

$\rm :\longmapsto\: {\bigg(\dfrac{k}{3} \bigg) }^{2} - 11\bigg(\dfrac{k}{3} \bigg) + k = 0 - - (3)$

$\rm :\longmapsto\:\dfrac{ {k}^{2} }{9} - \dfrac{11k}{3} + k = 0$

$\rm :\longmapsto\:\dfrac{ {k}^{2} - 33k + 9k}{9} = 0$

$\rm :\longmapsto\:\dfrac{ {k}^{2} -24k}{9} = 0$

$\rm :\longmapsto\:k(k - 24) = 0$

$\bf\implies \:k = 0 \: \: \: \: or \: \: \: \: 24$

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:Hence-\begin{cases} &\bf{k = 0} \\ &\bf{k = 24} \end{cases}\end{gathered}\end{gathered}$

Justification :-

Case :- 1

When k = 0,

The two equations reduces to

$\rm :\longmapsto\: {x}^{2} - 11x = 0$

$\rm :\longmapsto\: {x}(x - 11) = 0$

$\bf\implies \:x = 0 \: \: or \: \: 11$

and

$\rm :\longmapsto\: {x}^{2} - 14x = 0$

$\rm :\longmapsto\:x(x - 14) = 0$

$\bf\implies \:x = 0 \: \: or \: \: 14$

Hence,

$\red{ \boxed{ \sf{ \: 0 \: is \: the \: common \: root}}}$

Case :- 2

When k = 24

The two equations reduces to

$\rm :\longmapsto\: {x}^{2} - 11x + 24 = 0$

$\rm :\longmapsto\:(x - 8)(x - 3) = 0$

$\bf\implies \:x = 8 \: \: or \: \: 3$

and

$\rm :\longmapsto\: {x}^{2} - 14x + 48 = 0$

$\rm :\longmapsto\:(x - 8)(x - 6) = 0$

$\bf\implies \:x = 8 \: \: or \: \: 6$

Hence,

$\red{ \boxed{ \sf{ \: 8 \: is \: the \: common \: root}}}$

Additional Information :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

• If Discriminant, D > 0, then roots of the equation are real and unequal.

• If Discriminant, D = 0, then roots of the equation are real and equal.

• If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

• Discriminant, D = b² - 4ac