Practice Questions for Algebra - Quadratic Equations

1. PE 4 - Quadratic Equation | Algebra - Quadratic Equations

If 3x³ – 9x² + kx –12 is divisible by x – 3, then it is also divisible by:

2. PE 4 - Quadratic Equation | Algebra - Quadratic Equations

When ${(1 - \frac{1}{a})}^{6}$ is expanded, the sum of the last three coefficients is:

Answer: Option C

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Explanation :

When ${(1 - \frac{1}{a})}^{6}$ is expanded, the the last three coefficients are:

⁶C₄, −⁶C₅ and ⁶C₆

⁶C₄ = 15
−⁶C₅ = −6
⁶C₆ = 1

∴ The sum of coefficients = 15 – 6 + 1 = 10

Hence, option (c).

3. PE 4 - Quadratic Equation | Algebra - Quadratic Equations

For what value(s) of k does the pair of equations y = x² and y = 3x + k have two identical solutions?

4. PE 4 - Quadratic Equation | Algebra - Quadratic Equations

A man born in the first half of the nineteenth century was x years old in the year x². He was born in:

5. PE 4 - Quadratic Equation | Algebra - Quadratic Equations

The values of y which will satisfy the equations
2x² + 6x + 5y + 1 = 0
2x + y + 3 = 0
may be found by solving:

Answer: Option C

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Explanation :

2x² + 6x + 5y + 1 = 0
∴ 2x(x + 3) + 5y + 1 = 0 ...(1)

Also, 2x + y + 3 = 0
∴ 2x = – (y + 3) ...(2)
∴ x = (-(y+3))/2 ...(3)

Substituting (2) & (3) in (1),
-(y + 3) $[\frac{- (y + 3)}{2} + 3]$ + 5y + 1 = 0

∴ $\frac{- (y + 3) (- y - 3 + 6)}{2}$ + 5y + 1 = 0

∴ –(y + 3) (–y + 3) + 10y + 2 = 0

∴ –(9 – y²) + 10y + 2 = 0

∴ y² + 10y – 7 = 0

Hence, option (c).

6. PE 4 - Quadratic Equation | Algebra - Quadratic Equations

The sum of the roots of equation 4x² + 5 – 8x = 0 is equal to:

7. PE 4 - Quadratic Equation | Algebra - Quadratic Equations

The solution of $\sqrt{5 x - 1}$ + $\sqrt{x - 1}$ = 2 is:

8. PE 4 - Quadratic Equation | Algebra - Quadratic Equations

The sum of all the roots of 4x³ – 8x² – 63x – 9 = 0 is:

PE 4 - Quadratic Equation | Algebra - Quadratic Equations