Free Quadratic Equations 01 Practice Test - 10th Grade 

 Step 1:-   For $x^2–2x–3=0$, value of discriminant
                 $D=(-2{)}^2–4(-3\left)\right(1)=4+12=16$.

 Step 2:-  Since, D is a perfect square, roots are rational and unequal.

 Step 3:- Solving the equation,
$x=\frac{-(-2)+\sqrt{(}16)}{2}or\frac{-(-2)-\sqrt{(}16)}{2}$
$x=3,-1$
Thus, both of them are integers.

Given, the roots of quadratic equation $9x^2+6kx+4=0$ are real and equal. 

$\text{Discriminant,}\Delta =b^2-4ac\Delta =36k^2–4\left(4\right)\left(9\right)=36k^2–4\left(36\right)$

The roots of quadratic equation are real and equal then $\Delta =036k^2–4\left(36\right)=0k^2-4=0k^2=4k=\sqrt{4}\therefore k=\pm 2$

$\frac{x+5}{1}=\frac{2x+10}{x-6}$  
$\Rightarrow$(x + 5)(x - 6) = 2x + 10
$\Rightarrow x^2-6x+5x-30=2x+10$
$\Rightarrow x^2-3x-40=0$  is of the form of  $a{x}^2+bx+c$

Hence, is a quadratic equation.

Let, Length = $x$ 
Breadth = $y$

Given,
Area of Rectangle = $634{\text{m}}^2$
Length : $x$
Thrice the breadth : $3y$
2 more that thrice : $3y+2$
So, $x=3y+2$

$\text{Area of the rectangle}=\text{length}\times \text{breadth}$
$634=xy634=(2+3y)y634=2y+3y^2\therefore 3y^2+2y-634=0$

This equation resembles the general form of quadratic equation $a{x}^2+bx+c=0.$

Lets find the values of $y$ satisfying the equation.(Roots of the equation)

$y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-2\pm \sqrt{2^2-4\times 3(-634)}}{2\times 3}y=\frac{-2\pm \sqrt{4+7608}}{6}=\frac{-2\pm \sqrt{7612}}{6}y=\frac{-2\pm 87.246}{6}y=\frac{-2+87.246}{6}\text{or}\frac{-2-87.246}{6}y=14.20\text{or}-14.87$

Length is always positive.
$\therefore y=14.20\text{m}$

$x=2+3y\therefore x=2+3\left(14.20\right)=44.6\text{m}$

Length $=44.6\text{m}$ 
Breadth $=14.2\text{m}$

Let the numbers be $x\&x+1$

Square of consecutive number: $x^2\&(x+1{)}^2$
Sum of square: $x^2+(x+1{)}^2$
$\therefore x^2+(x+1{)}^2=365$

Lets solve
$x^2+x^2+2x+1=3652x^2+2x-364=0x^2+x-182=0$
(Dividing by 2)

Factorise,
$x^2+14x-13x-182=0x(x+14)-13(x+14)=0(x+14)(x-13)=0\therefore x=13\&-14$ 

Since numbers are positive integers, $x=13$

So, required consecutive numbers are $13\&14.$

By using the method of factorisation,

$x^2-3x-10=0$  
$\Rightarrow$ $x^2-5x+2x-10=0$       

$\Rightarrow x(x-5)+2(x-5)=0$       

$\Rightarrow$ $(x-5)(x+2)=0$       

$\Rightarrow x=5$ and $x=-2$

Multiply the equation by 5.

we get $25x^2–30x–10=0$

$(5x{)}^2–[2\times \left(5x\right)\times 3]+3^2–3^2–10=0$
$(5x–3{)}^2–9–10=0(5x–3{)}^2–19=0(5x–3{)}^2=195x-3=\pm \sqrt{19}x=\frac{3\pm \sqrt{19}}{5}$

Roots of the equation are $\frac{3+\sqrt{19}}{5}\&\frac{3-\sqrt{19}}{5}$

Let the number be $x$.
Square of number: $x^2$
Lesser by 15: $x^2-15$
of two times the number: $x^2-15=2x$

We get, $x^2-15=2x$
Lets solve, $x^2-2x-15=0x^2-5x+3x-15=0x(x-5)+3(x-5)=0(x-5)(x+3)=0\therefore x=5\&x=-3$

The number satisfying the given statement are $5\&-3$.

Given: $x=\frac{2}{3}$ is a solution of the quadratic equation $7x^2+mx-3=0$.

Substituting $x=\frac{2}{3}$ in the given quadratic equation we get

$\Rightarrow$  $7(\frac{2}{3}{)}^2+m(\frac{2}{3})-3=0$

$\Rightarrow$  $7\left(\frac{4}{9}\right)+m\left(\frac{2}{3}\right)-3=0$

$\Rightarrow$  $\frac{28}{9}+\frac{2m}{3}-3=0$

$\Rightarrow$  $28+6m-27=0$

$\Rightarrow$  $m=\frac{-1}{6}$

Step 1 : For, $x^2+2x+m=0$,
 $\Rightarrow a=1,b=2,c=m$

We know that

$D=b^2–4ac$
 $\Rightarrow D=(2{)}^2–4m$
          $=4-4m$     

Step 2 : The roots of a quadratic equation are real and distinct only when $D>0$

Step 3 :  $4-4m>0$ 

             $4>4m$

            $m<1$