Shortcut Methods
Typical numerical on Quadratic Equaltions while preparing for JEE & CBSE board exams
- SOLUTION: $$x^2 - 5x + 6 = 0$$
Here, $$(x - 2)(x - 3) = 0$$
$$ \therefore x = 2, 3$$
Ans: Roots = 2, 3
- SOLUTION: $$x^2 - 4x - 21 = 0$$
Here, $$(x + 3)(x - 7) = 0$$
$$\therefore x = - 3, 7$$
Ans: Roots = 7, -3
- SOLUTION: $$x^2 + kx - 5 = 0$$
By comparing the given equation with the standard quadratic equation $$ax^2 + bx + c = 0$$ we get: $$a = 1, b = k, \ \text{and} \ c = -5$$
Now, for equal roots, we must have: $$b^2 - 4ac = 0$$ $$\Rightarrow k^2 - 4 \times 1\times (-5) = 0 $$ $$\Rightarrow k^2 + 20 = 0 $$ $$\Rightarrow k^2 = -20$$
Ans: No reel solution
- SOLUTION: Given equations: $$x^2 + px + q = 0$$
Comparing with the standard quadratic equation $$ax^2 + bx + c = 0,$$ we get $$a = 1, b = p, \ \text{and }c = q$$
Since, $$\alpha + \beta = 10,$$ $$ \alpha\beta = 24$$
Thus applying Vieta’s relation, we have $$\alpha + \beta = \frac{-b}{a} \Rightarrow 10 = \frac{-p}{1}$$ $$\Rightarrow p = -10$$
$$\alpha\beta = \frac{c}{a} \Rightarrow 24 = \frac{q}{1}$$ $$q= 24$$
Ans: p = -10 and q = 24
- SOLUTION: $$(x - 1)(x + 3) = 12$$ $$x^2 + 3x - x- 3 = 12$$ $$x^2 + 2x - 3 -12=0$$ $$x^2 +2x - 15 = 0$$
Here, $$(x + 5)(x - 3)= 0$$
$$ \therefore x= -5, 3$$
Ans: Roots = 3 , -5
- SOLUTION: $$2x^2 + mx + 1 = 0$$ $$D = b^2 - 4ac$$ $$= m^2 - 4 \times 2 \times 1$$ $$= m^2 - 8$$
for reel and distinct roots D>0 i.e. $$m^2 - 8 > 0$$ $$(m - 2\sqrt{2})(m + 2\sqrt{2}) > 0$$ $$\therefore m > 2\sqrt{2} \ or \ m < -2\sqrt{2}$$
Ans: $$ m>2\sqrt2$$
- SOLUTION: Given points are: $$(1,4), (2,11), (3,20)$$
Let the equation of the parabola be $$y = ax^2 + bx + c$$
Substituting the values of the points, we get 4 = a + b + c ….. (i) 11 = 4a + 2b + c ….. (ii) 20 = 9a + 3b + c ….. (iii)
Subtracting (i) from (ii), we get $$7 = 3a + b$$ $$3a + b = 7….. (iv)$$
Subtracting (ii) from (iii), we get 9 = 5a + b $$5a+ b = 9 ….. (v)$$
Subtracting (v) from (iv), we get $$2a= 2$$ $$\Rightarrow a =1$$
Substituting the value of a in (iv) we get
$$3a+b=7$$ $$3 \times 1+ b=7$$ $$3+b =7$$ $$\therefore b =4$$
Substituting the values of a and b in (i) we get 4 = a + b + c $$4 = 1 +4 + c$$ $$c=-1$$
$$\therefore $$ $$Equation of the parabola:$$ y = x^2 + 4x -1$$
- SOLUTION: Given equations: $$(a - b)x^2 + (b - c)x + (c - a) = 0$$
Comparing with the standard quadratic equation $$ax^2 + bx + c = 0$$ we get: $$a = a - b, b = b - c, \ \text{and }c = c - a$$
Sum of roots: $$\alpha + \beta = \frac{-b}{a} = \frac{- (b - c)}{a - b} = \frac{c - b}{b-a}$$ $$= \frac{(a-b)-(c-b)}{a-b} $$ $$= \frac{a-c}{a-b}$$
Ans: $$a - c$$
- SOLUTION: Given equation: $$(p + q)x^2 + (q + r)x + (r + p) = 0$$
Comparing with the standard quadratic equation we get: $$a = p + q, b = q + r, a \ \text{and }c = r + p$$
Product of Roots: $$\alpha\beta = \frac{c}{a} = \frac{r + p}{p + q}$$
Ans: $$(r+p)/(p+q)$$
- SOLUTION: $$x^4 - 2x^2 + 1 = 0$$
Substituting $$x^2 = y$$ we get $$y^2 - 2y + 1 = 0$$
This is a standard quadratic equation.
$$\therefore y = \frac{2 \pm \sqrt{4 - 4 \times 1 \times 1}}{2 \times 1}$$ $$= \frac{2 \pm 0 }{2}$$ $$= 1$$
Substituting $$y = x^2$$ we get $$x^2 = 1$$ $$ \therefore x=\pm 1$$
Ans: $$x=\pm 1$$
