Question

if -4 is the root of the equation x2+2x+4p=0 find values of k for which the equation x3+p(1+3k) x+7(3+4k)=0 has equal roots

Answer 1

Answer:

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Step-by-step explanation:

x²+2x+4p=0

(-4)²+2(-4)+4p=0

16-8+4p=0

p=-2

x³+p(1+3k)x+7(3+4k)=0

(-4)³+(-2)(1+3k)(-4)+7(3+4k)=0

-64-(2-6k)-4+21+28k=0

-64+8+24k+21+28k=0

-56+21+52k=0

-35=-52k

k=35/52

Answer 2

Given that,

The root of the equation = -4

The equation is

$x^2+2x+4p=0$...(I)

The other equation is

$x^2+p(1+3k)x+7(3+4k)=0$

We need to find the value of p

Using equation (I)

$x^2+2x+4p=0$

Put the value of x=-4

$(-4)^2+2\times(-4)+4p=0$

$16-8+4p=0$

$4p=-8$

$p=-2$

We need to calculate the value of k

Using equation (II)

$x^2+p(1+3k)x+7(3+4k)=0$

This equation has equal roots.

So, D = 0

$(p(1+3k))^2-4\times1\times7(3+2k)=0$

Put the value of p

$((-2)^2(1+3k))^2-4\times1\times7(3+2k)=0$

$4(1+9k^2+6k)-28(3+2k)=0$

$4(1+9k^2+6k-21-14k)=0$

$9k^2-8k-20=0$

$9k^2-18k+10k-20=0$

$9k(k-2)+10(k-2)=0$

$(9k+10)(k-2)=0$

$k=2, k=\dfrac{-10}{9}$

Hence, The value of k is 2 and $\dfrac{-10}{9}$