Free Quadratic Equations 03 Practice Test - 10th Grade 

If one root of the equation $2x^2+ax+6=0$ is 3, then substituting $x=3$ will satisfy the equation.

$2(3{)}^2+3a+6=0$

 $\Rightarrow 18+3a+6=0$

 $\Rightarrow 24+3a=0$

  $\Rightarrow a=-8$

Step 1: For $2x^2–6x+8=0$, value of discriminant
            $D=(-6{)}^2–4(2\left)\right(8)=36-64=-28$

Step 2: Since $D<0$, roots are imaginary.

Step 1:- For, $x^2+2(k+2)x+9k=0$, value of discriminant $D=\left[2\right(k+2){]}^2–4(9k)=4(k^2+4-5k)$

Step 2:- The roots of quadratic equation are real and equal only when $D=0$
$k^2+4-5k=0$
$\Rightarrow k^2-5k+4=0$
$\Rightarrow k^2-k-4k+4=0$
$\Rightarrow k(k-1)-4(k-1)=0$
$\Rightarrow (k-1)(k-4)=0$

Step 3:-   $k=4or1$ 

${2x}^2-7x+3=0$
Dividing by the coefficient of $x^2$, we get
$x^2-\frac{7}{2}x+\frac{3}{2}=0;a=1,b=\frac{7}{2},c=\frac{3}{2}$

Adding and subtracting the square of $\frac{b}{2}=\frac{7}{4}$, (half of coefficient of x)

we get,
$[x^2-2(\frac{7}{4})x+(\frac{7}{4}{)}^2]-(\frac{7}{4}{)}^2+\frac{3}{2}=0$ 

The equation after completing the square is :
$(x-\frac{7}{4}{)}^2-\frac{25}{16}=0$

Taking square root, $(x-\frac{7}{4})=(\pm \frac{5}{4})$
Taking positive sign $\frac{5}{4},x=3$
Taking negative sign $\frac{-5}{4},x=\frac{1}{2}$

Step 1:-  $x^2+2(k+2)x+9=0$,  $\Rightarrow a=1,b=2(k+2),c=9$

                  $D=4(k+2{)}^2–4(9)=4(k^2+4k-5)$       [$D=b^2-4ac$ ]

Step 2:- The roots of quadratic equation are real and equal only when $D=0$

                 $4(k^2+4k-5)=0,\Rightarrow k^2+4k-5=0$

Step 3:-    $k^2+4k-5=0$

                 $\Rightarrow k^2+5k-k-5=0$

                  $\Rightarrow k(k+5)-1(k+5)=0$

                  $\Rightarrow (k-1)(k+5)=0$

                   $\Rightarrow k=1ork=-5$

Given: A can do a piece of work in $x$ days and B in $x+16$ days.
Work done by A in one day = $\frac{1}{x}$
Work done by B in one day = $\frac{1}{x+16}$
Work done by A and B together in one day = $\frac{1}{15}$
$\Rightarrow$   $\frac{1}{x}$ +  $\frac{1}{x+16}$ =  $\frac{1}{15}$
$\Rightarrow$   $\frac{2x+16}{x(x+16)}=\frac{1}{15}$
$\Rightarrow$   $x^2+16x=15(2x+16)$
$\Rightarrow$   $x^2-14x-240=0$
$\Rightarrow$   $x^2-24x+10x-240=0$
$\Rightarrow$   $x(x-24)+10(x-24)=0$
$\Rightarrow$   $(x-24)(x+10)=0$
$\Rightarrow$   $x=24orx=-10$
$\Rightarrow$   $x=24$ as $x$ cannot be negative.

Given:  $4(2x+3{)}^2-(2x+3)-14=0$
Substitute $(2x+3)=y$, Hence the given equation reduces to
$4y^2-y-14=0$
$\Rightarrow$  $4y^2-8y+7y-14=0$
$\Rightarrow$  $4y(y-2)+7(y-2)=0$
$\Rightarrow$  $(4y+7)(y-2)=0$
$\Rightarrow$  $y=\frac{-7}{4}ory=2$
When  $y=-\frac{7}{4}$, 
$(2x+3)=-\frac{7}{4}$
$2x=-\frac{19}{4}$
$\Rightarrow$  $x=-\frac{19}{8}$
When  $y=2$, 
$(2x+3)=2$
$2x=-1$
$\Rightarrow$  $x=-\frac{1}{2}$

Since, x = 2 is a root of the given equation $2x^2+ax-6=0$

$\to$ $2(2{)}^2+a\times 2-6=0$

                  $8+2a-6=0$

                  $2a=-2$

                  $a=-1$

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

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

$\Rightarrow$  $x(x+1)=72$ 

$\Rightarrow$  $x^2+x-72=0$

$\Rightarrow$  $x^2+9x-8x-72=0$

$\Rightarrow$   $x(x+9)-8(x+9)=0$

$\Rightarrow$   $(x-8)(x+9)=0$

$\Rightarrow$   $x=8orx=-9$

Since the given numbers are natural numbers, we get $x=8$

Hence the two consecutive natural numbers are 8 and 9.