For the Quadratic equation $ax^2 +bx +c =0$ Where a, b and c are real numbers and a ≠0 Roots of the quadratic equation is given by Quadratic Formula
$x= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
Question 1. State which all quadratic equations have real roots, no real roots
a. $x^2 + 8x+7=0$
b. $3x^2 +5x+1=0$
c. $9x^2 +x -3=0$
d. $11x^2 +5x+1=0$
e. $10x^2 +3x+7=0$
f. $2x^2 -6x+3=0$
g. $2x^2 -x-2=0$ Solution Nature of roots of Quadratic equation
S.no
Condition
Nature of roots
1
$b^2 -4ac > 0$
Two distinct real roots
2
$b^2-4ac =0$
One real root
3
$b^2-4ac < 0$
No real roots
Real roots: :(a), (b), (c),(d) ,(e),(f),(g)
No real roots : (d) ,(e)
Question 2. Find the roots of the quadratic equation using Quadratic Formula
a. $x^2-3x-10=0$
b. $x^2 -11x+30=0$ Solution
a.
$x^2-3x-10=0$
Comparing this to $ax^2 + bx +c=0$
We have a =1, b=-3 and c =-10
$x= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
or
$x=\frac {3 \pm \sqrt {9 +40}}{2}$
$x=\frac {3 \pm 7}{2}$
or
$x=\frac {3 + 7}{2}$ or $x=\frac {3 - 7}{2}$
So roots are x=5 and -2
b. $x^2 -11x+30=0$
Comparing this to $ax^2 + bx +c=0$
a=1, b=-11 and c =30
$x= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
$x=\frac {11 \pm \sqrt {121 -120}}{2}$
$x=\frac {11 \pm 1}{2}$
or
$x=\frac {11 + 1}{2}$ or $x=\frac {11 - 1}{2}$
So
Roots are 5 and 6
Question 3. Find the roots of the quadratic equation using Quadratic formula
a. $x^2 +4x-5=0$
b. $2x^2-7x+3=0$
Solution
a.
$x^2 +4x-5=0$
Comparing this to $ax^2 + bx +c=0$
a=1, b=4 and c =-5
$x= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
So
x=1 or -5
b.
$2x^2-7x+3=0$
Comparing this to $ax^2 + bx +c=0$
a=2, b=-7 and c =3
$x= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
So
x=1/2 or 3
Question 4 Find the roots using Quadratic Formula
a. $x^2+ (7 - x)^2= 25$
b. $y^2+ (y+2)^2 =580$
c. $11x^2 -31x -6 =0$
d. $9 -y - 10y^2 =0$
e. $14x +4x^2 =2x -5$
f. $3y^2 + 4y=2(y+4)$
g. $2x^2 -5x +3 =0$
h. $\frac {x+1}{x-1} + \frac {x-2}{x+2} =3$ ,$x \neq 1$, $x \neq -2$
i. $x^2 + 5x =-1$
j. $ \sqrt {2x + 9} + x=13$
k. $\frac {1}{a} + \frac {1}{b} + \frac {1}{x} = \frac {1}{a+b+x}$
l. $ \sqrt {2} x^2 + 7x + 5 \sqrt {2}=0$
m. $ \frac {3}{x+1} + \frac {4}{x-1} = \frac {29}{4x-1};x \neq 1,-1,1/4$ Solution
a.
$x^2+ (7 - x)^2= 25$
$x^2 + 49 +x^2 -14x =25$
$2x^2 -14x + 24=0$
$x^2 -7x + 12=0$
Comparing this to $ax^2 + bx +c=0$
a=1, b=-7 and c =12
$x= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
So
x=3 or 4
b.
$y^2+ (y+2)^2 =580$
$y^2 + y^2 + 4 + 4y =580$
$2y^2 + 4y -576=0$
$y^2 + 2y-288=0$
Comparing this to $ay^2 + by +c=0$
a=1, b=2 and c =-288
$y= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
So
y=16 or -18
c.
$11x^2 -31x -6 =0$
Comparing this to $ax^2 + bx +c=0$
a=11, b=-31 and c =-6
$x= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
So
x=3 or -2/11
d.
$9 -y - 10y^2 =0$
$10y^2 + y -9=0$
Comparing this to $ay^2 + by +c=0$
a=10, b=1 and c =-9
$y= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
So
y=-1 or 9/10
e.
$14x +4x^2 =2x -5$
$4x^2 + 12x +5=0$
Comparing this to $ax^2 + bx +c=0$
a=4, b=12 and c =5
$x= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
so x=-1/2 or -5/2
f.
$3y^2 + 4y=2(y+4)$
$3y^2 + 4y = 2y + 8$
$3y^2 + 2y -8=0$
Comparing this to $ay^2 + by +c=0$
a=3, b=2 and c =-8
$y= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
so y=-2 or 4/3
g.
$2x^2 -5x +3 =0$
Comparing this to $ax^2 + bx +c=0$
a=2, b=-5 and c =3
$x= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
so y=3/2 or 1
h.
$\frac {x+1}{x-1} + \frac {x-2}{x+2} =3$
$ \frac { (x+1)(x+2) + (x-2)(x-1)}{(x-1)(x+2)}=3$
$ x^2 + 3x +2 + x^2 -3x + 2= 3(x^2 +x -2)$
$2x^2 +4 = 3x^2 +3x -6$
$x^2 +3x -10=0$
Comparing this to $ax^2 + bx +c=0$
a=1, b=3 and c =-10
$x= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
so x=-5 or 2
i.
$x^2 + 5x =-1$
$x^2 + 5x +1=0$
Comparing this to $ax^2 + bx +c=0$
a=1, b=5 and c =1
$x= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
so,$x = \frac {-5 + \sqrt {21}}{2}$ or $x= \frac {-5 - \sqrt {21}}{2}$
