Question

| Q196) If the quadratic equations 2x2+4x+(a+5)=0
have equal roots and (a+4)x2+ax-3b=0 have
distinct real roots then which of the following is
true:
1) a = -3,b<3 2) a = 3, b>
16
3) a = -3, b> 3
Olt
4) a = 3, b < 3
Olt​

Answer 1

Answer:

1) a = -3, b < 3

Step-by-step explanation:

Given,

$2x^2 + 4x + (a+5)=0$ has equal root,

Since, if a quadratic equation ax² + bx + c = 0 has equal root,

Then,

$b^2 = 4ac$

And, if it has distinct real roots,

Then, $b^2 > 4ac$

So,

$4^2 = 4\times 2\times (a+5)$

$16=8(a+5)$

$2=a+5$

$\implies a = -3$

$\implies (a+4)x^2 + ax - 3b = 0\implies x^2 - 3x - 3b = 0$

$(-3)^2 > 2\times 1\times -3b$

$9 > -6b$

$-2 > b$

$\implies b < 3$

Hence, OPTION 1) is correct.

Answer 2

The value of a=-3 and  $b>-\frac{3}{4}$

Step-by-step explanation:

In a quadratic equation $ax^2+bx+c=0$

Root are equal then $b^2=4ac$

Roots are distinct then $b^2>4ac$

We have given,

$2x^2 + 4x + (a+5)=0$ has equal root.

i.e. on comparing we get,

$4^2 = 4\times 2\times (a+5)$

$16=8(a+5)$

$2=a+5$

$a=-3$

Now substitute 'a' in $(a+4)x^2+ax-3b=0$

$(-3+4)x^2+(-3)x-3b=0$

$x^2 - 3x - 3b = 0$

We have given,  $x^2 - 3x - 3b = 0$ have distinct roots.

i.e. $(-3)^2 > 4\times 1\times -3b$

$9 > -12b$

$\frac{9}{12}>-b$

$\frac{3}{4}>-b$

$b>-\frac{3}{4}$

The value of a=-3 and  $b>-\frac{3}{4}$

#Learn more

Show that the equation $x^2+ax-4=0$ has real and distinct roots.

https://brainly.in/question/5202739