If the equations k 6 x 2-3- rx -2 x 2-1-0 and 6 k 2 x 2-1- px -4 x 2-2-0- k -0 have both roots common- then the valueof p - r isA- 2B- 1-2C- 1D- 4

The correct option is A 2(6k+2)x^2+rx+3k 1=0 ⋯ (1) (12k+4)x^2+px+6k 2=0 ⋯ (2) Since the roots of eqns. (1) and (2) are equal, 6k+212k+4=rp=3k 16k 2 ∴ ...

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Byju's Answer

Standard XIII

Mathematics

Common Roots

If the equati...

Question

If the equations $k (6 x^{2} + 3) + r x + (2 x^{2} - 1) = 0$ and $6 k (2 x^{2} + 1) + p x + (4 x^{2} - 2) = 0, k \neq 0$ have both roots common, then the value of $\frac{p}{r}$ is

2

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Solution

The correct option is A $2$
$(6 k + 2) x^{2} + r x + 3 k - 1 = 0 \dots (1)$
$(12 k + 4) x^{2} + p x + 6 k - 2 = 0 \dots (2)$
Since the roots of eqns. $(1)$ and $(2)$ are equal,
$\frac{6 k + 2}{12 k + 4} = \frac{r}{p} = \frac{3 k - 1}{6 k - 2} ∴ \frac{p}{r} = 2$

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Similar questions

Q. lf the equations

k (6 x^{2} + 3) + r x + (2 x^{2} - 1) = 0

and

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

have both roots common, then the value

\frac{p}{r}

Q. If both the roots of the equations

k (6 x^{2} + 3) + r x + 2 x^{2} - 1 = 0

and

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

are common, then

2 r - p

is equal to-

Q. If the equation

k (6 x^{2} + 3) + r x + (2 x^{2} - 1) = 0

and

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

have symmetrical roots, then

2 r - p

equals

Q. If the equations

k (6 x^{2} + 3) + r x + 2 x^{2} - 1 = 0 a n d 6 k (2 x^{2} + 1) + p x + 4 x^{2} - 2 = 0

have both the roots common, find the value of

2 r - p

If both the roots of $k (6 x^{2} + 3) + r x + 2 x^{2} - 1 = 0$ and $2 (6 k + 2) x^{2} + p x + 2 (3 k - 1) = 0$ are common, then 2r-p is equal to

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If the equations k(6x2+3)+rx+(2x2−1)=0 and 6k(2x2+1)+px+(4x2−2)=0,k≠0 have both roots common, then the value of pr is

The correct option is A 2(6k+2)x2+rx+3k−1=0 ⋯(1) (12k+4)x2+px+6k−2=0 ⋯(2) Since the roots of eqns. (1) and (2) are equal, 6k+212k+4=rp=3k−16k−2∴pr=2

If the equations $k (6 x^{2} + 3) + r x + (2 x^{2} - 1) = 0$ and $6 k (2 x^{2} + 1) + p x + (4 x^{2} - 2) = 0, k \neq 0$ have both roots common, then the value of $\frac{p}{r}$ is

The correct option is A $2$
$(6 k + 2) x^{2} + r x + 3 k - 1 = 0 \dots (1)$
$(12 k + 4) x^{2} + p x + 6 k - 2 = 0 \dots (2)$
Since the roots of eqns. $(1)$ and $(2)$ are equal,
$\frac{6 k + 2}{12 k + 4} = \frac{r}{p} = \frac{3 k - 1}{6 k - 2} ∴ \frac{p}{r} = 2$