25. Find the value(s) of k so that the equations x² - 11x + k = 0 and

1.	25. Find the value(s) of k so that the equations x² - 11x + k = 0 and x² - 14x + 2k = 0 may havea common root.⚠️ No spam ⚠️
Answer» $\large\underline{\sf{Solution-}}$ 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

25. Find the value(s) of k so that the equations x² - 11x + k = 0 and x² - 14x + 2k = 0 may havea common root.⚠️ No spam ⚠️

Answer»

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

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.

25. Find the value(s) of k so that the equations x² - 11x + k = 0 and x² - 14x + 2k = 0 may havea common root.⚠️ No spam ⚠️

Now