## GRE: Quadratic Equation

If you set the polynomial $$a{ x }^{ 2 }+bx+c$$ equal to zero, where a, b and c are constants and

a ≠ 0 , there is a special name for it. It is called a quadratic, You can find the values for x that make the equation true.

Example:     $${ x }^{ 2 }-3x+2=0$$

To find the solutions, also called roots of an equation, start by factoring whenever possible. You can factor $${ x }^{ 2 }-3x+2$$ into (x-2)  (x-1) , making the quadratic equation:

(x-2) (x-1) = 0

To find the roots, set each binomial equal to 0. That gives you:-

(x-2) = 0      or     (x-1) =0

Solving for x, you get x=2 or x=1

The solutions to a quadratic in the form $$a{ x }^{ 2 }+bx+c=0$$ can also be found using the quadratic formula provided a, b and c are real numbers and a ≠ 0 then:

$$x\quad =\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$$

EXAMPLE:- Find the Solution of 2x2 + 9x +9 = 0

Solution :- a=2 , b=9 , c=9

$$x\quad =\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$$ $$x\quad =\quad \frac { -9\pm \sqrt { 81-4(2)(9) } }{ 2(2) }$$

=  (-9±3) / 4

ANS= -3     or -3/2

Sum of the roots of a quadratic equation =  -b/a

Product of the roots of a quadratic equation = c/a

• $$D\quad =\quad { b }^{ 2 }-4ac$$ is the discriminant of a quadratic equation.

If D < 0 (i.e. the discriminant is negative) then the equation has no real roots.

If D= 0 then the quadratic equation has two equal roots.

If D> 0 (i.e. the discriminant is positive) then the quadratic equation has two distinct roots.

Graph of a Quadratic Expression:-

Let f(x)= ax2 + bx +c , Where a, b, c are real and a ≠ 0, Then y=f(x) represents a parabola, Whose axis is parallel to y- axis. This gives the following cases:

1) a> 0 and D = (b2 -4ac) < 0 (Roots are imaginary).

$$f(x)>0\quad \forall \quad x\in R$$

2) a > 0 and D = 0 (The roots are real and equal)

f(x) will be positive for all values of x except at the vertex where f(x) = 0

3) When a> 0 and D > 0 (The roots are real and distinct)

F(x) will be equal to zero when x is equal to either of $$\alpha \quad or\quad \beta$$.

$$f(x)>0\quad \forall x\in (-\infty ,\alpha )\bigcup (\beta ,\infty )\\ and\quad f(x)<0\quad \forall x\in (\alpha ,\beta )$$

4) When a < 0 and D=0 (Roots are imaginary)

$$f(x)<0\quad \forall x\in R$$

5) When a<0 and D=0 (Roots are real and equal)

F(x) is negative for all values of x except at the vertex Where f(x)= 0

6) When a<0 and D>0 (Roots are real and distinct)

$$f(x)<0\quad \forall x\in (-\infty ,\quad \alpha )\bigcup (\beta ,\quad \infty )\\ f(x)>0\quad \forall x\in (\alpha ,\quad \beta )$$

## Sample Problems

Problem 1)  If -4 is a solution for x in the equation $${ x }^{ 2 }+kx+8=0$$, what is k?

Sol.  If -4 is a solution for x in the equation.

Then,$${ (-4) }^{ 2 }+k(-4)+8=0\\ 16-4k+8=0\\ -4k=-24\\ k={ (6) }^{ Ans }$$

Problem 2) Given that $${ x }^{ 2 }-13x=30$$, what is x?

Sol.

$${ x }^{ 2 }-13x=30\\ { x }^{ 2 }-13x-30=0\\ x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } \\ =(\frac { 13\pm \sqrt { 169-4(-30) } }{ 2 } )\\ =\frac { 13\pm \sqrt { 289 } }{ 2 } =\frac { 13\pm 17 }{ 2 } \\ =\frac { -4 }{ 2 } \quad or\quad \frac { 30 }{ 2 } \\ =-2\quad or\quad 15$$

Problem 3) If the area of a certain square (expressed in square meters) is added to its perimeter (expressed in meters), the sum is 77. What is the length of a side of the square?

Sol.  Let side of square be x.

Area if square $$={ x }^{ 2 }$$

Perimeter of square = 4x

$${ x }^{ 2 }+4x=77\\ { x }^{ 2 }+4x-77=0\\ \alpha ,\beta =\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2 } \\ \alpha =\frac { -4+\sqrt { 16+4(77) } }{ 2 } ,\quad \beta =\frac { -4-\sqrt { 16+4(77) } }{ 2 } \\ \alpha =\frac { -4+18 }{ 2 } ,\quad \beta =\frac { -4-18 }{ 2 } \\ \alpha =7,\quad \beta =-11$$

Since the ride of a square must be positive, negative value (-11) has been discorded.

Side of the square = 7

Problem 4)  If $${ x }^{ 2 }-6x-27=0$$ and $${ y }^{ 2 }-6y-4=0$$, what is the maximum value of x+y?

Sol.

$${ x }^{ 2 }-6x-27=0\\ \alpha =\frac { -b+\sqrt { { b }^{ 2 }-4ac } }{ 2a } ,\quad \beta =\frac { -b-\sqrt { { b }^{ 2 }-4ac } }{ 2a } \\ \alpha =\frac { 6+\sqrt { 36-4(-27) } }{ 2 } ,\quad \beta =\frac { 6-\sqrt { 36-4(-27) } }{ 2 } \\ \alpha =\frac { 6+\sqrt { 144 } }{ 2 } ,\quad \beta =\frac { 6-\sqrt { 144 } }{ 2 } \\ \alpha =\frac { 6+12 }{ 2 } ,\quad \beta =\frac { 6-12 }{ 2 } \\ \alpha =9,\quad \beta =-3$$

x may be 9 or -3.

$${ y }^{ 2 }-6y-40=0\\ \alpha =\frac { -b+\sqrt { { b }^{ 2 }-4ac } }{ 2a } ,\quad \beta =\frac { -b-\sqrt { { b }^{ 2 }-4ac } }{ 2a } \\ \alpha =\frac { 6+\sqrt { 36-4(-40) } }{ 2 } ,\quad \beta =\frac { 6-\sqrt { 36-4(-40) } }{ 2 } \\ \alpha =\frac { 6+\sqrt { 196 } }{ 2 } ,\quad \beta =\frac { 6-\sqrt { 196 } }{ 2 } \\ \alpha =\frac { 6+14 }{ 2 } ,\quad \beta =\frac { 6-14 }{ 2 } \\ \alpha =10,\quad \beta =-4$$

y may be 10 or -4.

For maximum value of (x+y),

x should be maximum & y should be maximum.

Maximum value of (x+y) = 9+10 = 19

Problem 5)  xy>0

Quantity A: $${ (x+y) }^{ 2 }$$

Quantity B: $${ (x-y) }^{ 2 }$$

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

d) The relationship cannot be determined from the information given.

Sol.  Quantity A: $${ (x+y) }^{ 2 }$$

Quantity B: $${ (x-y) }^{ 2 }$$

Now subtract $${ x }^{ 2 }+{ y }^{ 2 }$$ from both columns.

Then, Quantity A:       2xy

Quantity B:         -2xy

xy>0

xy is positive, so the value in quantity A will be positive, regardless of the values of x and y. Similarly, the value in Quantity B will always be negative, regardless of the values of x & y.

Quantity A is larger.

Problem 6)  Find the value of $$\sqrt { 6+\sqrt { 6+\sqrt { 6+——— } } }$$

a) 4

b) 3

c) 5

d) 5

Sol.  Let

$$y\quad =\quad \sqrt { 6+\sqrt { 6+\sqrt { 6+——— } } } \\ y\quad =\quad \sqrt { 6+y } \quad (Squaring\quad both\quad sides)\\ { y }^{ 2 }=6+y\\ { y }^{ 2 }-y-6=0\\ \alpha =\frac { 1+\sqrt { 1-4(-6) } }{ 2 } ,\quad \beta =\frac { 1-\sqrt { 1-4(-6) } }{ 2 } \\ \alpha =\frac { 1+5 }{ 2 } ,\quad \beta =\frac { 1-5 }{ 2 } \\ \alpha =3,\quad \beta =-2\\ Therefore,\quad y=3\quad (Since\quad \sqrt { 6+\sqrt { 6+\sqrt { 6—— } } } cannot\quad be\quad negative)$$

Problem 7)  If the roots of the equation , differ by 2, then which of the following is true?

a) $${ c }^{ 2 }=4(c+1)$$

b) $${ b }^{ 2 }=4(c+1)$$

c) $${ c }^{ 2 }=b+4$$

d) $${ b }^{ 2 }=4(c+2)$$

Sol.

$${ x }^{ 2 }-bx+c=0$$

Let $$\alpha \quad and\quad \beta$$ be the roots of equation.

$$\alpha +\beta =b\quad \quad \alpha \beta =c\\ \alpha -\beta =2\\ { (\alpha +\beta ) }^{ 2 }={ \alpha }^{ 2 }+{ \beta }^{ 2 }+2\alpha \beta \\ (Subtract\quad 4\alpha \beta \quad from\quad both\quad sides)\\ { (\alpha +\beta ) }^{ 2 }-4\alpha \beta ={ \alpha }^{ 2 }+{ \beta }^{ 2 }+2\alpha \beta -4\alpha \beta \\ { \alpha }^{ 2 }-2\alpha \beta +{ \beta }^{ 2 }\\ ={ (\alpha -\beta ) }^{ 2 }\\ \\ { b }^{ 2 }-4c={ 2 }^{ 2 }=4\\ { b }^{ 2 }=4c+4\\ Therefore,\quad { ({ b }^{ 2 }=4(c+1)) }^{ Ans }$$