If the following quadratic equation has two e | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Comparing the given quadratic equation $x^{2}-(k+10)x+9(k+1)=0$ with the standard form $ax^{2}+bx+c=0$,we get:$a=1, b=-(k+10), c=9(k+1)$.Since the

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) Comparing the given quadratic equation $x^{2}-(k+10)x+9(k+1)=0$ with the standard form $ax^{2}+bx+c=0$,we get:$a=1, b=-(k+10), c=9(k+1)$.Since the quadratic equation has two equal and real roots,the discriminant $D$ must be equal to $0$.$D = b^{2}-4ac = 0$Substituting the values:$( -(k+10) )^{2} - 4(1)(9(k+1)) = 0$$(k+10)^{2} - 36(k+1) = 0$$k^{2} + 20k + 100 - 36k - 36 = 0$$k^{2} - 16k + 64 = 0$This is a perfect square trinomial:$(k-8)^{2} = 0$Taking the square root on both sides:$k-8 = 0$$k = 8$Therefore,the value of $k$ is $8$.

If the following quadratic equation has two equal and real roots,then find the value of $k$: $x^{2}-(k+10)x+9(k+1)=0$.

Similar Questions

Solve the following equation using the quadratic formula,if the equation has a solution in $R$: $x^{2}+6x+5=0$.

Examine whether the following equation is quadratic or not: $(x-3)(2x+1) = x(x+5)$

Find whether the following equation has real roots. If real roots exist,find them.
$5x^{2}-2x-10=0$

If one of the roots of the equation $kx^{2} + 2x - 3 = 0$ is $2$,then find $k$.

The equation $9x^2 + 30x + 25 = 0$ has .............. in $R$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module