RD Sharma Test: Quadratic Equations - Class 10 MCQ

x = 3 satisfies only the equation x²−5x+6 = 0
As (3)²−5×3+6 = 0
⇒ 9−15+6 = 0
⇒ 15−15 = 0
⇒ 0 = 0
⇒ L.H.S. = R.H.S.

Given: p (x) = 2x² + x - 6 = 0 and q (x) = x² - 3x — 10 = 0
∴ p (-2) = 2(-2)² + (-2) - 6 = 0 = 8 - 2 - 6 = 8 - 8 = 0
∴ q (-2) = (-2)² - 3 (-2) - 10 = 0 = 4 + 6 - 10 = 10 - 10 = 0
Since p (-2) = 0 and q (-2) = 0 therefore, - 2 is the common root of 2x² + x - 6 = 0 and x² - 3x - 10 = 0 

Let the number of wickets taken by Srinath be x then the number of wickets taken by Kumble will be 2x−3

According to question, x (2x−3) = 20

⇒ 2x²−3x−20 = 0

⇒ 2x²−8x+5x−20 = 0

⇒ 2x(x−4)+5(x−4) = 0

⇒ (x−4)(2x+5) = 0

⇒ x−4 = 0and 2x+5 = 0

Therefore, number of wickets taken by Srinath is 4.
Then number of wickets taken by Kumble = 2 x 4 - 3 = 5 

Given: (x²+1)² − x² = 0
⇒ x² + 1  + 2x − x² = 0
⇒ 2x + 1 = 0
⇒ x = −1/2
Therefore, (x²+1)² − x² = 0 has no real roots.

Explanation here:a²x²−3abx+2b² = 0
⇒ a²x²−2abx−abx+2b² = 0
⇒ ax(ax−2b)−b(ax−2b) = 0
⇒ (ax−b)(ax−2b) = 0
⇒ x−b = 0and ax−2b = 0

Let one multiple of 5 be x then the next consecutive multiple of will be (x+5) According to question,

x = −35 is not possible therefore x = 30
Then the other multiple of 5 is x+5 = 30+5 = 35
Then the number are 30 and 35.

Since b² - 4ac < 0 therefore, x² - 4x + 3√2 = 0 has no real roots 

In equation x²+x−5 = 0
a = 1,b = 1,c = −5
∴ b²−4ac = (1)²−4×1×(−5) = 1 + 20 = 21
Since b²−4ac > 0 therefore, x²+x−5 = 0 has two distict roots.

Since sum of the roots = -b/a ∴ a+ b = -a/1 = -a

Given: p(x) = 3x²−px−2 = 0
∴ p(2) = 3(2)²−p(2)−2 = 0
⇒ 12−2p−2 = 0
⇒ −2p = −10
⇒ p = 5

Let Sharma’s present age be x years then his age 3 years ago is (x−3) years and 5 years from now is (x+5) years. According to question,

But x = 3 does not satisfy the given condition.
Therefore, Sharma's present age is 7 years. 

If ax²+bx+c = 0 has equal roots, then
b²−4ac = 0
⇒ 4ac = b²
⇒ c = 

If the quadratic equation 6x2 - 2kx + 6 = 0 has equal roots, then

Let one root be α then other root will be 3α.
∴ Sum of the roots

And Product of the roots 

Equating the coefficients of α² and subtracing e.q. (ii) from e.q. (i), we get

Given: 5x²+8x+4 = 2x²+4x+6
⇒ 5x²−2x²+8x−4x+4−6 = 0
⇒ 3x²+4x−2 = 0
Here, the degree is 2, therefore it is a quadratic equation.

Given: 3√3x²+10x+√3 = 0 Discriminant = b²−4ac = (10)²−4×3√3×√3 = 100−36 = 64