1 Crore+ students have signed up on EduRev. Have you? 
...(i)
Squaring both sides of (1), we get
Since x cannot be negative, therefore, neglect
Thus, x = 2
The roots of ax^{2} + bx + c = 0, a ≠ 0 are real and unequal, if b^{2} – 4ac is _______.
Given equation is ax^{2} + bx + c = 0
Roots are real and unequal, if b^{2} – 4ac > 0
If one root of the equation a ( b – c ) x^{2} + b(c – a)x + c(a – b) = 0 is 1, then the other root is ___.
Given equation is a(b – c)x^{2} + b(c – a)x + c(a – b) = 0
Let α be the other root, then
Product of roots =
If 2 is a root of the equation x^{2} + bx + 12 = 0 and the equation x^{2} + bx + q = 0 has equal roots, then q is equal to
Since x = 2 is a root of the equation
x^{2} + bx + 12 = 0 ⇒ (2)^{2} + b(2) + 12 = 0
⇒ 2b = –16 ⇒ b = –8
Then, the equation x^{2} + bx + q becomes
x^{2 }– 8x + q = 0 ...(1)
Since (1) has equal roots ⇒ b^{2} – 4ac = 0
⇒ (–8)^{2} – 4(1)q = 0 ⇒ q = 16
One of the two students, while solving a quadratic equation in x, copied the constant term incorrectly and got the roots 3 and 2. The other copied the constant term and coefficient of x^{2} correctly as –6 and 1 respectively. The correct roots are ____.
Let the equation be x^{2} + ax + b = 0
Its roots are 3 and 2
∴ Sum of roots, 5 = –a
and product of roots, 6 = b
∴ Equation is x^{2} – 5x + 6 = 0
Now constant term is wrong and it is given that correct constant term is – 6.
∴ x^{2} – 5x – 6 = 0 is the correct equation.
Its roots are –1 and 6.
If the roots of the equation (a – b)x^{2} + (b – c)x + (c – a) = 0 are equal. Then _______.
Since the given equation has equal roots.
∴ D = 0
⇒ (b – c)^{2} – 4(c – a) (a – b) = 0
⇒ 4a^{2} + b^{2} + c^{2} – 4ab + 2bc – 4ac = 0
⇒ (–2a)^{2} + (b)^{2} + (c)^{2} + 2(–2a)(b) + 2bc + 2(–2a)(c) = 0
⇒ (–2a + b + c)^{2} = 0
⇒ –2a + b + c = 0 ⇒ 2a = b + c
If one of roots of 2x^{2} + ax + 32 = 0 is twice the other root, then the value of a is ________.
Equation is 2x^{2 }+ ax + 32 = 0
Let one root be α, then other would be 2α.
Now, α × 2α = 16 ⇒ α = ±2√2
and α + 2α = –a/2 ⇒ 3α = –a/2 ⇒ 6α = –a
⇒ ±12√2 = −a or a = ±12√2
Given equation is ...(1)
Rationalising (1), we get
...(2)
Adding (1) and (2), we get
Roots of the quadratic equation x^{2} + x – (a + 1)(a + 2) = 0 are ________.
Given equation is x^{2} + x – (a + 1)(a + 2) = 0
⇒ x^{2} + (a + 2)x – (a + 1)x – (a + 1) (a + 2) = 0
⇒ x(x + (a + 2)) – (a + 1)(x + (a + 2)) = 0
⇒ (x – (a + 1))(x + (a + 2)) = 0
⇒ x = (a + 1) or x = – (a + 2)
For what value of a, the roots of t he equation 2x^{2} + 6x + a = 0, satisfy the condition (where α and β are the roots of equation).
Given equation is 2x^{2} + 6x + a = 0
Now,
⇒
⇒
⇒ (9/2) < a
22 docs62 tests

Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 
22 docs62 tests







