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Standard XII

Mathematics

Conditions for a system of linear equations to have a unique solution

Solve for k...

Question

Solve for $k$ : $k (k - 1) (3 k + 2) = 0$

k = - 1, k = 1

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k = 0, k = 1

k = \frac{- 2}{3}

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k = 0, k = 0

k = \frac{- 2}{3}

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k = 0, k = 1, k = 2

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Solution

The correct option is A $k = 0, k = 1$ , $k = \frac{- 2}{3}$
Given, $k (k - 1) (3 k + 2) = 0$

Applying zero product rule, we get

$k = 0, k - 1 = 0, 3 k + 2 = 0$

$∴ k = 0, k = 1$ , $k = \frac{- 2}{3}$

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Similar questions

Q. Find the values of k for which the roots are real and equal in each of the following equations:

(i)

k x^{2} + 4 x + 1 = 0

(ii)

k x^{2} - 2 \sqrt{5} x + 4 = 0

(iii)

3 x^{2} - 5 x + 2 k = 0

(iv)

4 x^{2} + k x + 9 = 0

(v)

2 k x^{2} - 40 x + 25 = 0

(vi)

9 x^{2} - 24 x + k = 0

(vii)

4 x^{2} - 3 k x + 1 = 0

(viii)

x^{2} - 2 (5 + 2 k) x + 3 (7 + 10 k) = 0

(ix)

(3 k + 1) x^{2} + 2 (k + 1) x + k = 0

(x)

k x^{2} + k x + 1 = - 4 x^{2} - x

(xi)

(k + 1) x^{2} + 2 (k + 3) x + (k + 8) = 0

(xii)

x^{2} - 2 k x + 7 k - 12 = 0

(xiii)

(k + 1) x^{2} - 2 (3 k + 1) x + 8 k + 1 = 0

(xiv)

(2 k + 1) x^{2} + 2 (k + 3) x + (k + 5) = 0

(xvii)

4 x^{2} - 2 (k + 1) x + (k + 4) = 0

(xviii)

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If $k_{1}, k_{2}, k_{3}$ are real numbers and the equation $k_{1} x^{2} + k_{2} x + k_{3} = 0$ have three roots, then value of $k_{1} + k_{2} + k_{3}$ is

Q. Let

C_{k} =^{n} C_{k}

for

0 \leq k \leq n

and

A_{k} = [\begin{matrix} C_{k - 1}^{2} & 0 0 & C_{k}^{2} \end{matrix}]

for

k \geq 1

and

A_{1} + A_{2} + . . . . A_{n} = [\begin{matrix} k_{1} & 0 0 & k_{2} \end{matrix}]

then

Find the values of k for which the roots are real and equal in each of the following equations.
$(i) k x^{2} + 4 x + 1 = 0 (i i) k x^{2} - 2 \sqrt{5} x + 4 = 0 (i i i) 3 x^{2} - 5 x + 2 k = 0 (i v) 4 x^{2} + k x + 9 = 0 (v) 2 k x^{2} - 40 x + 25 = 0 (v i) 9 x^{2} - 24 x + k = 0 (v i i) 4 x^{2} - 2 k x + 1 = 0 (v i i i) x^{2} - 2 (5 + 2 k) x + 3 (7 + 10 k) = 0 (i x) (3 k + 1) x^{2} + 2 (k + 1) x + k = 0 (x) k x^{2} + k x + 1 = - 4 x^{2} - x (x i) (k + 1) x^{2} + 2 (k + 3) x (k + 8) = 0 (x i i) x^{2} - 2 k x + 7 k - 12 = 0 (x i i i) (k + 1) x^{2} - 2 (3 k + 1) x + 8 k + 1 = 0 (x i v) 5 x^{2} - 4 x + 2 + k (4 x^{2} 1 - 2 x - 1) = 0 (x v) (4 - k) x^{2} + (2 k + 4) x + (8 k + 1) = 0 (x v i) (2 k + 1) x^{2} + 2 (k + 3) x + (k + 5) = 0 (x v i i) 4 x^{2} - 2 (k + 1) x + (k + 4) = 0 (x v i i i) 4 x^{2} - 2 (k + 1) x + (k + 1) = 0$

Q. Find the values of k for which the following equations have real and equal roots:
(i)

x^{2} - 2 (k + 1) x + k^{2} = 0

(ii)

k^{2} x^{2} - 2 (2 k - 1) x + 4 = 0

(iii)

(k + 1) x^{2} - 2 (k - 1) x + 1 = 0

(iv)

x^{2} + k (2 x + k - 1) + 2 = 0

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Solve for k: k(k−1)(3k+2)=0

The correct option is A k=0,k=1, k=−23Given, k(k−1)(3k+2)=0Applying zero product rule, we getk=0,k−1=0,3k+2=0∴k=0,k=1, k=−23

Solve for $k$ : $k (k - 1) (3 k + 2) = 0$

The correct option is A $k = 0, k = 1$ , $k = \frac{- 2}{3}$
Given, $k (k - 1) (3 k + 2) = 0$

Applying zero product rule, we get

$k = 0, k - 1 = 0, 3 k + 2 = 0$

$∴ k = 0, k = 1$ , $k = \frac{- 2}{3}$