$(x+1)^{2}-x^{2}$=0
$X^{2}+2x+1-x^{2} $= 0
2x+1=0
x=-$\dfrac{1}{2}$
Hence, there is one real root.
Given, $100x^{2} – 20x + 1$=0
$100x^{2} – 10x – 10x + 1$ = 0
10x(10x – 1) -1(10x – 1) = 0
$(10x – 1)^{2}$ = 0
∴ (10x – 1) = 0 or (10x – 1) = 0
⇒x =$ \dfrac{1}{10}$ or x = $\dfrac{1}{10}$
Let x is one number
Another number = 27 – x
Product of two numbers = 182
x(27 – x) = 182
⇒ $x^{2} – 27x – 182$ = 0
⇒$ x^{2} – 13x – 14x + 182$ = 0
⇒ x(x – 13) -14(x – 13) = 0
⇒ (x – 13)(x -14) = 0
⇒ x = 13 or x = 14
Given x=$\dfrac{1}{2}$ as root of equation $x^{2}-mx-\dfrac{5}{4}$=0.
($\dfrac{1}{2} )^{2} – m(\dfrac{1}{2} ) – \dfrac{5}{4}$ = 0
$\dfrac{1}{4} -\dfrac{m}{2}-\dfrac{5}{4}$ =0
m=-2
Let the base be x cm.
Altitude = (x – 7) cm
In a right triangle,
$Base^{2} + Altitude^{2} $= $Hypotenuse^{2} $(From Pythagoras theorem)
∴$ x^{2} + (x – 7)^{2}$ = $13^{2}$
By solving the above equation, we get;
⇒ x = 12 or x = – 5
Since the side of the triangle cannot be negative.
Therefore, base = 12cm and altitude = 12-7 = 5cm
$ 2x^{2} + x + 4$ = 0
⇒ $ 2x^{2} + x$ = -4
Dividing the equation by 2, we get
⇒ $ x^{2} + \dfrac{1}{2}x$ = -2
⇒ $ x^{2} + 2 × x × \dfrac{1}{4}$ = -2
By adding ( $ \dfrac{1}{4})^{2}$ to both sides of the equation, we get
⇒ $ (x)^{2} + 2 × x × \dfrac{1}{4} + ( \dfrac{1}{4})^{2}$
= $ ( \dfrac{1}{4})^{2} – 2$
⇒$ (x + \dfrac{1}{4})^{2} $ = $ \dfrac{1}{16}– 2$
⇒ $ (x + \dfrac{1}{4})^{2}$ = $ \dfrac{-31}{16}$
The square root of negative number is imaginary, therefore, there is no real root for the given equation.
If one root is reciprocal of others, then the product of roots will be:
$\alpha \times \dfrac{1}{\alpha}$ =$\dfrac{ (k-4)}{4}$
k-4=4
k=8
A quadratic equation $ax^{2} + bx + c $= 0 has no real roots, if $b^{2} – 4ac$ < 0.
That means, the quadratic equation contains imaginary roots.
Let x and (x + 1) be the two consecutive integers.
According to the given,
x(x + 1) = 360
$ x^{2} + x $ = 360
$ x^{2} + x – 360$
$(x + 1)^{2} – 2(x + 1)$ = 0
$x^{2}+ 1 + 2x – 2x – 2$= 0
$x^{2} – 1$ = 0
$x^{2} $= 1
x = ± 1