Question

By calculating ,let us write the value/values of K for which each of the following Quadratic equations has real and equal roots ...


I) X²—2(5+2k)x+3(7+10k)=0
ii) (3k+1)X²+2(k+1)x+k=0 ​

Answer 1

Answer:

11\26

Step-by-step explanation:

for real and equal roots D=0 also D=b square- 4ac putting the value we get

for 1 equation

x2-2(5 +2k)x+3(7+10k)=0

here a=1 b=10+4k and c=21+30k

putting the value we get

D=10+4k-(21+30k)

D=10-21+4k-30k

D=-11-26k

-11=-26k

so k=-11\-26

k =11\26

hope this help you

Answer 2

$\bf{\Huge{\boxed{\bf{\red{ANSWER\::}}}}}$

$\bf{Given\begin{cases}\sf{x^{2} -2(5+2k)x+3(7+10k)=0}\\ \sf{(3k+1)x^{2} +2(k+1)x+k=0}\end{cases}}$

$\bf{\Large{\underline{\bf{To\:find\::}}}}}$

The value of k.

$\bf{\Large{\underline{\rm{\blue{Explanation\::}}}}}$

We know that b² - 4ac = 0 the roots of the quadratic are real and equal, then in case;

$\leadsto\sf{Each\:root\:\:=\:\frac{-b+0}{2a} =\frac{-b}{2a} .}$

The expression D = b² - 4ac is called discriminate.

We can compared this given equation are;

Ax² + Bx + C = 0

So,

• A = 1
• B = -2(5+2k)
• C = 3(7+10k)

$\bf{\Large{\boxed{\tt{\orange{Using\:formula\::}}}}}}$

$\longmapsto\sf{b^{2} -4ac=0}$

$\longmapsto\sf{[-2(5+2k)]^{2} -4*1*3(7+10k)=0}$

$\longmapsto\sf{(-10-4k)^{2} -12(7+10k)=0}$

$\longmapsto\sf{(-10)^{2} +(4k)^{2} -2*-10*-4k-84-120k=0}$

$\longmapsto\sf{100+16k^{2} -2*(-40k)-84-120k=0}$

$\longmapsto\sf100+16k^{2} +80k-84-120k=0}$

$\longmapsto\sf{16k^{2} -40k+16=0}$

$\longmapsto\sf{8(2k^{2} -5k+2)=0}$

$\longmapsto\sf{2k^{2}-5k+2=\cancel{\frac{0}{8}} }$

$\longmapsto\sf{2k^{2} -5k+2=0}$

$\longmapsto\sf{2k^{2} -4k-k+2=0}$

$\longmapsto\sf{2k(k-2)-1(k-2)=0}$

$\longmapsto\sf{(k-2)=0\:\:\:or\:\:\:(2k-1)=0}$

$\longmapsto\sf{k=2\:\:\:or\:\:\:2k=1}$

$\longmapsto\sf{\red{k=2\:\:\:or\:\:\:k=\frac{1}{2}} }$

_______________________________________________

• A = (3k+1)
• B = 2(k+1)
• C = k

Therefore,

$\longmapsto\sf{b^{2} -4ac=0}$

$\longmapsto\sf{[2(k+1)]^{2} -4*(3k+1)*k=0}$

$\longmapsto\sf{(2k+2)^{2} -4k*(3k+1)=0}$

$\longmapsto\sf{(2k)^{2} +(2)^{2} +2*2k*2-12k^{2} -4k=0}$

$\longmapsto\sf{4k^{2} +4+8k-12k^{2} -4k=0}$

$\longmapsto\sf{-8k^{2} +4k+4=0}$

$\longmapsto\sf{-4(2k^{2} -k-1)=0}$

$\longmapsto\sf{2k^{2} -k-1=\cancel{\frac{0}{-4}} }$

$\longmapsto\sf{2k^{2} -k-1=0}$

$\longmapsto\sf{2k^{2} -2k+k-1=0}$

$\longmapsto\sf{2k(k-1)+1(k-1)=0}$

$\longmapsto\sf{(k-1)=0\:\:\:or\:\:\:(2k+1)=0}$

$\longmapsto\sf{k=1\:\:\:or\:\:\:2k=-1}$

$\longmapsto\sf{\red{k=1\:\:\:or\:\:\:k=-\frac{1}{2}} }$