Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

The roots of the equation $x^{2} + 6 x - (a^{2} + 2 a - 8) = 0$ are [[1]] and [[2]].

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is A - 2, - A + 4.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

The roots of the equation

x^{2} + 6 x - (a^{2} + 2 a - 8) = 0

are

a - 2

and

- (a + 4)

.
Given quadratic equation is,

x^{2} + 6 x - (a^{2} + 2 a - 8) = 0

.
We know that,
Discriminant for the quadratic equation

a x^{2} + bx + c = 0, a \neq 0

where a, b and c are real numbers is given by,

D = b^{2} - 4 ac

.
On comparing the given equation

x^{2} + 6 x - (a^{2} + 2 a - 8) = 0

with standard form we get,

a = 1, b = 6, c = - (a^{2} + 2 a - 8)

Then,

D = b^{2} - 4 ac

⟹ D = 6^{2} - 4 \times 1 \times \{- (a^{2} + 2 a - 8)\}

⟹ D = 36 + {4 a}^{2} - 8 a - 32

⟹ D = {4 a}^{2} - 8 a + 4

⟹ D = {4 (a + 1)}^{2} > 0

Discriminant is greater than zero which means the roots of the equation exist.
Since, quadratic formula for the quadratic equation

a x^{2} + bx + c = 0, a \neq 0

where a, b and c are real numbers is given by,

x = \frac{- b \pm \sqrt{D}}{2 a}

Then,

x = \frac{- 6 \pm \sqrt{{4 (a + 1)}^{2}}}{2 \times 1}

⟹ x = \frac{- 6 \pm 2 (a + 1)}{2}

⟹ x = - 3 \pm (a + 1)

On taking positive sign,

⟹ x = - 3 + (a + 1)

⟹ x = a - 2

On taking negative sign,

⟹ x = - 3 - (a + 1)

⟹ x = - a - 4 = - (a + 4)

Therefore, the roots of the equation

x^{2} + 6 x - (a^{2} + 2 a - 8) = 0

are

a - 2

and

- (a + 4)

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!