Quantitative Aptitude

Algebra

3,715 Questions
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Algebra Questions

Multiple choice

If $b_1b_2=2(c_1+c_2)$, then at least one of the equations $x^2+b_1x+c_1=0$ and $x^2+b_2x+c_2=0$ has?

  1. Real roots

  2. Purely imaginary roots

  3. Imaginary roots

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The discriminant of the first equation is D1 = b1^2 - 4c1 and the second is D2 = b2^2 - 4c2. Summing them: D1 + D2 = b1^2 + b2^2 - 4(c1 + c2). Given b1*b2 = 2(c1 + c2), we have D1 + D2 = b1^2 + b2^2 - 2(b1*b2) = (b1 - b2)^2. Since the sum of discriminants is a square, it is >= 0, implying at least one discriminant must be >= 0, meaning at least one equation has real roots.

Multiple choice

Let $a,b,c$ be distinct non-negative numbers and the vectors $a\widehat{i}+a\widehat{j}+c\widehat{k},\widehat{i}+\widehat{k},c\widehat{i}+c\widehat{j}+b\widehat{k}$ lie in the plane, then the quadratic equation $ax^2+2cx+b=0$ has

  1. real and equal roots

  2. real and unequal roots

  3. both roots real and positive

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The condition that the three vectors lie in the same plane means their scalar triple product is zero. Calculating the determinant of the matrix formed by these vectors leads to the relation c^2 = ab. Substituting this into the quadratic equation ax^2 + 2cx + b = 0, the discriminant becomes (2c)^2 - 4ab = 4c^2 - 4ab = 4(ab) - 4ab = 0, which implies real and equal roots.

Multiple choice

If $6$ lies between the roots of the equation $x ^ { 2 } + 2 ( a - 3 ) x + 9 = 0$ then
  1. $a \in \left( - \dfrac { 3 } { 4 } , \infty \right)$
  2. $a \in \left( - \infty , - \dfrac { 3 } { 4 } \right)$
  3. $a \in ( - \infty , 0 ) \cup ( 6 , \infty )$
  4. $a \in \left( - \dfrac { 3 } { 4 } , 6 \right)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a quadratic f(x) = x^2 + 2(a-3)x + 9, if 6 lies between the roots, then f(6) < 0. f(6) = 36 + 12(a-3) + 9 = 36 + 12a - 36 + 9 = 12a + 9. Setting 12a + 9 < 0 gives 12a < -9, so a < -3/4.

Multiple choice

The equation $\displaystyle x^{2} +ax+b^{2 }=0$ has two roots each of which exceeds a number $c$, then
  1. $\displaystyle a^{2} < 4b^{2}$
  2. $\displaystyle c^{2} +ac+b^{2 }> 0$
  3. $\displaystyle -\dfrac{a}{2}< c$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a quadratic f(x) = x^2 + ax + b^2 to have roots > c, the vertex -a/2 must be > c, f(c) > 0, and the discriminant D = a^2 - 4b^2 >= 0. Option B, c^2 + ac + b^2 > 0, is equivalent to f(c) > 0, which is a necessary condition for the roots to be greater than c.

Multiple choice

If both the roots of a quadratic equation $x ^ { 2 } - 6 a x + \left(2 - 2 a + 9 a ^ { 2 }\right) = 0$ exceed $3$ then
  1. $a > \dfrac { 9 } { 11 }$
  2. $a \geq \dfrac { 11 } { 9 }$
  3. $a > \dfrac { 11 } { 9 }$
  4. $a < \dfrac { 11 } { 9 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For roots of x^2 - 6ax + (2-2a+9a^2) = 0 to exceed 3: 1) Discriminant >= 0: 36a^2 - 4(2-2a+9a^2) >= 0 => 36a^2 - 8 + 8a - 36a^2 >= 0 => 8a >= 8 => a >= 1. 2) Vertex > 3: 6a/2 > 3 => 3a > 3 => a > 1. 3) f(3) > 0: 9 - 18a + 2 - 2a + 9a^2 > 0 => 9a^2 - 20a + 11 > 0. Roots of 9a^2 - 20a + 11 are 1 and 11/9. Inequality holds for a < 1 or a > 11/9. Combining conditions, a > 11/9.