Problems on Quadratic Equation

We will solve different types of problems on quadratic equation using quadratic formula and by method of completing the squares. We know the general form of the quadratic equation i.e., ax$^{2}$ + bx + c = 0, that will help us to find the nature of the roots and formation of the quadratic equation whose roots are given.

1. Solve the quadratic equation 3x$^{2}$ + 6x + 2 = 0 using quadratic formula.

Solution:

The given quadratic equation is 3x$^{2}$ + 6x + 2 = 0.

Now comparing the given quadratic equation with the general form of the quadratic equation ax$^{2}$ + bx + c = 0 we get,

a = 3, b = 6 and c = 2

Therefore, x = $\frac{- b ± \sqrt{b^{2} - 4ac}}{2a}$

⇒ x = $\frac{- 6 ± \sqrt{6^{2} - 4(3)(2)}}{2(3)}$

⇒ x = $\frac{- 6 ± \sqrt{36 - 24}}{6}$

⇒ x = $\frac{- 6 ± \sqrt{12}}{6}$

⇒ x = $\frac{- 6 ± 2\sqrt{3}}{6}$

⇒ x = $\frac{- 3 ± \sqrt{3}}{3}$

Hence, the given quadratic equation has two and only two roots.

The roots are $\frac{- 3 - \sqrt{3}}{3}$ and $\frac{- 3 - \sqrt{3}}{3}$.

2. Solve the equation 2x$^{2}$ - 5x + 2 = 0 by the method of completing the squares.

 Solutions:

The given quadratic equation is 2x$^{2}$ - 5x + 2 = 0

Now dividing both sides by 2 we get,

x$^{2}$ - $\frac{5}{2}$x + 1 = 0

⇒ x$^{2}$ - $\frac{5}{2}$x = -1

Now adding $(\frac{1}{2} \times \frac{-5}{2})$ = $\frac{25}{16}$ on both the sides, we get

⇒ x$^{2}$ - $\frac{5}{2}$x + $\frac{25}{16}$ = -1 + $\frac{25}{16}$

⇒ $(x - \frac{5}{4})^{2}$ = $\frac{9}{16}$

⇒ $(x - \frac{5}{4})^{2}$ = ($\frac{3}{4}$)$^{2}$

⇒ x - $\frac{5}{4}$ = ± $\frac{3}{4}$

⇒ x = $\frac{5}{4}$ ± $\frac{3}{4}$

⇒ x = $\frac{5}{4}$ - $\frac{3}{4}$ and $\frac{5}{4}$ + $\frac{3}{4}$

⇒ x = $\frac{2}{4}$ and $\frac{8}{4}$

⇒ x = $\frac{1}{2}$ and 2

Therefore, the roots of the given equation are $\frac{1}{2}$ and 2.

3. Discuss the nature of the roots of the quadratic equation 4x$^{2}$ - 4√3 + 3 = 0.

Solution:

The given quadratic equation is 4x$^{2}$ - 4√3 + 3 = 0

Here the coefficients are real.

The discriminant D = b$^{2}$ - 4ac = (-4√3 )$^{2}$ - 4 ∙ 4 ∙ 3 = 48 - 48 = 0

Hence the roots of the given equation are real and equal.

4. The coefficient of x in the equation x$^{2}$ + px + q = 0 was taken as 17 in place of 13 and thus its roots were found to be -2 and -15. Find the roots of the original equation.

Solution:

According to the problem -2 and -15 are the roots of the equation x$^{2}$ + 17x + q = 0.

Therefore, the product of the roots = (-2)(-15) = $\frac{q}{1}$

⇒ q = 30.

Hence, the original equation is x$^{2}$ – 13x + 30 = 0

⇒ (x + 10)(x + 3) = 0

⇒ x = -3, -10

Therefore, the roots of the original equation are -3 and -10.

11 and 12 Grade Math 

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