Typical numerical on Quadratic Equaltions while preparing for JEE & CBSE board exams

1. SOLUTION: $$x^2-5x+6=0$$

Here, $$(x-2)(x-3)=0$$

$$\therefore x=2,3$$

Ans: Roots = 2, 3

2. SOLUTION: $$x^2-4x-21=0$$

Here, $$(x+3)(x-7)=0$$

$$\therefore x=-3,7$$

Ans: Roots = 7, -3

3. SOLUTION: $$x^2+kx-5=0$$

By comparing the given equation with the standard quadratic equation $$a{x}^2+bx+c=0$$ 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

4. SOLUTION: Given equations: $$x^2+px+q=0$$

Comparing with the standard quadratic equation $$a{x}^2+bx+c=0,$$ 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

5. 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

6. 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. $$m^2-8>0$$ $$(m-2\sqrt{2})(m+2\sqrt{2})>0$$ $$\therefore m>2\sqrt{2}orm<-2\sqrt{2}$$

Ans: $$m>2\sqrt{2}$$

7. SOLUTION: Given points are: $$(1,4),(2,11),(3,20)$$

Let the equation of the parabola be $$y=a{x}^2+bx+c$$

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 $$7=3a+b$$ $$3a+b=7\dots ..(iv)$$

Subtracting (ii) from (iii), we get 9 = 5a + b $$5a+b=9\dots ..(v)$$

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$$

8. SOLUTION: Given equations: $$(a-b)x^2+(b-c)x+(c-a)=0$$

Comparing with the standard quadratic equation $$a{x}^2+bx+c=0$$ 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: $$a-c$$

9. 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: $$(r+p)/(p+q)$$

10. 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: $$x=\pm 1$$