Free Quadratic Equations 02 Practice Test - 10th Grade 

On comparing $x^2–4x+k=0$ with standard form $a{x}^2+bx+c=0$, we get

a = 1, b = -4 and c = k

Now, discriminant, D = $b^2-4ac$
$\Rightarrow D=(-4{)}^2–4(1)k=16-4k$

The roots of quadratic equation are co-incident only when $D=0$.

 $\Rightarrow 16-4k=0$

 $\Rightarrow k=4$

Let the smaller number be $x$. Then the larger number will be $x+1$.

Their product is $x(x+1)$.

$x(x+1)=56$ 

$x^2+x-56=0$

$x^2+x+(\frac{1}{2}{)}^2–(\frac{1}{2}{)}^2-56=0$

$(x+\frac{1}{2}{)}^2-\frac{225}{4}=0$

$x+\frac{1}{2}=\frac{15}{2},-\frac{15}{2}$

$x=7,-8$  

The required number is $7$ as -8 is not a natural number.

We have, $x^2-3x+2=0$ 

$\Rightarrow D=b^2-4ac$

$\Rightarrow (-3{)}^2-4\times 1\times 2>0$

$\Rightarrow 1>0$ 

Since, $D>0$, roots are distinct and real.

Given Height = 6
$\Rightarrow -16t^2+24t+1=6$

$\Rightarrow 16t^2-24t+5=0$

$\Rightarrow 16t^2-4t-20t+5=0$

$\Rightarrow 16t^2-4t-20t+5=0$

$\Rightarrow 4t(4t-1)-5(4t-1)=0$

$\Rightarrow (4t-1)(4t-5)=0$

$t=\frac{1}{4},\frac{5}{4}or0.25,1.25$

So, at time  0.25 secs and 1.25 secs, the ball will be at a height of 6 feet.

Let the length of the legs of the triangle be $x$ and $x+4$  inches.

Applying Pythagoras' theorem,

$x^2+(x+4{)}^2=(20{)}^2{x}^2+x^2+8x+16=4002x^2+8x-384=0x^2+4x-192=0$

Solving the quadratic equation,$x^2+16x-12x-192=0x(x+16)-12(x+16)=0(x+16)(x-12)=0\therefore x=-16\&12$

Length can't be negative, hence
$x=12\text{inches}$

The shortest side(leg) is 12 .

Given, $x^2+4x+c=0$

Value of discriminant, $\Delta =b^2-4ac=4^2–4c=16-4c$

The roots of quadratic equation are real then $\Delta \ge 016–4c\ge 0-4c\ge -164c\le 16$
(The inequality changes when the equation is multiplied by ${}^{\prime }{-}^{ve}{}^{\prime }$

$\therefore c\le 4$

We have,

$9x^2-9(a+b)x+(2a^2+5ab+2b^2)=0$

Comparing this equation with $A{x}^2+Bx+C=0$, we have

$A=9,B=-9(a+b)andC=2a^2+5ab+2b^2$

$\therefore D=B^2-4AC$

$\Rightarrow D=81(a+b{)}^2-36(2a^2+5ab+2b^2)$

$\Rightarrow D=81(a^2+b^2+2ab)-(72a^2+180ab+72b^2)$

$\Rightarrow D=9a^2+9b^2-18ab$

$\Rightarrow D=9(a^2+b^2-2ab)$

$\Rightarrow D=9(a-b{)}^2\ge 0$

$\Rightarrow D\ge 0$

So, the roots of the given equation are real and are given by

$\alpha =\frac{-B+\sqrt{D}}{2A}=\frac{9(a+b)+3(a-b)}{18}=\frac{12a+6b}{18}=\frac{2a+b}{3}$

and, $\beta =\frac{-B-\sqrt{D}}{2A}=\frac{9(a+b)-3(a-b)}{18}=\frac{6a+12b}{18}=\frac{a+2b}{3}$

We know that roots of the quadratic equation is equal when D = 0
$\Rightarrow$$b^2-4ac=0$
So, from the equation x² − kx + 36 = 0
a = 1, b = -k and c = 36
$\Rightarrow$ $k^2-4\left(ac\right)=0$
     $k^2-4\left(36\right)=0$
     $k^2=144$ 
$\therefore$ k = 12 or -12

Let Peter's current age  = x years
$\therefore$ Age 13 years ago  =  $(x-13)$ years
Age 11 years later  = $(x+11)$ years

According to the given statement
11 years later , the age of Peter = Half the square of the age he was 13 years ago
$\Rightarrow$  $(x+11)$ = $\frac{(x-13{)}^2}{2}$
$\Rightarrow$  $2x+22=x^2-26x+169$
$\Rightarrow$ $x^2-28x+147=0$
$\Rightarrow$ $x^2-(21+7)x+147=0$
$\Rightarrow$ $x^2-21x-7x+147=0$
$\Rightarrow$ $x(x-21)-7(x-21)=0$
$\Rightarrow$ $x=21orx=7$

Since Peter's age can't be 7 years
$\therefore$ His current age = 21 years

Suppose B alone can takes$x$ days to finish the work.
Then A alone can finish it in $(x-6)$ days.
B's 1 day's work = $\frac{1}{x}$.
A's 1 day's work = $\frac{1}{x-6}$.
(A+B)'s 1 day's work = $\frac{1}{4}$
$\therefore \frac{1}{x}+\frac{1}{x-6}=\frac{1}{4}$
$\Rightarrow \frac{x-6+x}{x(x-6)}=\frac{1}{4}$
$\Rightarrow 8x-24=x^2-6x$
$\Rightarrow x^2-14x+24=0$
$\Rightarrow x^2-12x-2x+24=0$
$\Rightarrow (x-12)(x-2)=0$
$\Rightarrow x=12orx=2$
But $x$ cannot be less than 6.
$\Rightarrow x=12$
Hence, B alone can finish the work in 12 days