In the following, determine whether the given quadratic equations have real roots and if so, find the roots:

Solution:

In the following parts we have to find the real roots of the equations

(i) We have been given,

$16 x^{2}=24 x+1$

$16 x^{2}-24 x-1=0$

Now we also know that for an equation $a x^{2}+b x+c=0$, the discriminant is given by the following equation:

$D=b^{2}-4 a c$

Now, according to the equation given to us, we have, $a=16, b=-24$ and $c=-1$.

$D=b^{2}-4 a c$

$=576+64$

$=640$

Since, in order for a quadratic equation to have real roots,.Here we find that the equation satisfies this condition, hence it has real roots.

Now, the roots of an equation is given by the following equation,

$x=\frac{-b \pm \sqrt{D}}{2 a}$

Therefore, the roots of the equation are given as follows,

$x=\frac{-(-24) \pm \sqrt{640}}{2(16)}$

$=\frac{24 \pm 8 \sqrt{10}}{32}$

$=\frac{3 \pm \sqrt{10}}{4}$

Now we solve both cases for the two values of x. So, we have,

$x=\frac{3+\sqrt{10}}{4}$

Also,

$x=\frac{3-\sqrt{10}}{4}$

Therefore, the roots of the equation are $\frac{3+\sqrt{10}}{4}$ and $\frac{3-\sqrt{10}}{4}$

(ii) We have been given, $x^{2}+x+2=0$

Now we also know that for an equation $a x^{2}+b x+c=0$, the discriminant is given by the following equation:

$D=b^{2}-4 a c$

Now, according to the equation given to us, we have, $a=1, b=1$ and $c=2$.

Therefore, the discriminant is given as,

$D=(1)^{2}-4(1)(2)$

$=1-8$

$=-7$

Since, in order for a quadratic equation to have real roots,.Here we find that the equation does not satisfies this condition, hence it does not have real roots.

(iii) We have been given, $\sqrt{3} x^{2}+10 x-8 \sqrt{3}=0$

Now we also know that for an equation $a x^{2}+b x+c=0$, the discriminant is given by the following equation:

$D=b^{2}-4 a c$

Now, according to the equation given to us, we have, $a=\sqrt{3}, b=10$ and $c=-8 \sqrt{3}$.

Therefore, the discriminant is given as,

$D=(10)^{2}-4(\sqrt{3})(-8 \sqrt{3})$

$=100+96$

$=196$

Since, in order for a quadratic equation to have real roots,.Here we find that the equation satisfies this condition, hence it has real roots.

Now, the roots of an equation is given by the following equation,

$x=\frac{-b \pm \sqrt{D}}{2 a}$

Therefore, the roots of the equation are given as follows,

$x=\frac{-(10) \pm \sqrt{196}}{2(\sqrt{3})}$

$=\frac{-10 \pm 14}{2 \sqrt{3}}$

$=\frac{-5 \pm 7}{\sqrt{3}}$

Now we solve both cases for the two values of x. So, we have,

$x=\frac{-5+7}{\sqrt{3}}$

$=\frac{2}{\sqrt{3}}$

Also,

$x=\frac{-5-7}{\sqrt{3}}$

$=-4 \sqrt{3}$

Therefore, the roots of the equation are $\frac{2}{\sqrt{3}}$ and $-4 \sqrt{3}$.

(iv) We have been given, $3 x^{2}-2 x+2=0$

Now we also know that for an equation $a x^{2}+b x+c=0$, the discriminant is given by the following equation:

$D=b^{2}-4 a c$

Now, according to the equation given to us, we have, $a=3, b=-2$ and $c=2$.

$D=b^{2}-4 a c$

Now, according to the equation given to us, we have, $a=3, b=-2$ and $c=2$.

Therefore, the discriminant is given as,

$D=(-2)^{2}-4(3)(2)$

$=4-24$

 

$=-20$

Since, in order for a quadratic equation to have real roots,.Here we find that the equation does not satisfies this condition, hence it does not have real roots.

(v) We have been given, $2 x^{2}-2 \sqrt{6} x+3=0$

Now we also know that for an equation $a x^{2}+b x+c=0$, the discriminant is given by the following equation:

$D=b^{2}-4 a c$

Now, according to the equation given to us, we have, $a=2, b=-2 \sqrt{6}$ and $c=3$.

 

Therefore, the discriminant is given as,

$D=(-2 \sqrt{6})^{2}-4(2)(3)$

$=24-24$

 

$=0$

Since, in order for a quadratic equation to have real roots,.Here we find that the equation satisfies this condition, hence it has real and equal roots.

Now, the roots of an equation is given by the following equation,

$x=\frac{-b \pm \sqrt{D}}{2 a}$

Therefore, the roots of the equation are given as follows,

$x=\frac{-(2 \sqrt{6}) \pm 0}{2(2)}$

$=\frac{-\sqrt{6}}{2}$

$=-\sqrt{\frac{3}{2}}$

Therefore, the roots of the equation are real and equal and its value is $-\sqrt{\frac{3}{2}}$

(vi) We have been given, $3 a^{2} x^{2}+8 a b x+4 b^{2}=0$

Now we also know that for an equation $a x^{2}+b x+c=0$, the discriminant is given by the following equation:

$D=b^{2}-4 a c$

Now, according to the equation given to us, we have, $a=3 a^{2}, b=8 a b$ and $c=4 b^{2}$.

Therefore, the discriminant is given as,

$D=(8 a b)^{2}-4\left(3 a^{2}\right)\left(4 b^{2}\right)$

$=64 a^{2} b^{2}-48 a^{2} b^{2}$

 

$=16 a^{2} b^{2}$

Since, in order for a quadratic equation to have real roots,.Here we find that the equation satisfies this condition, hence it has real roots.

Now, the roots of an equation is given by the following equation,

$x=\frac{-b \pm \sqrt{D}}{2 a}$

Therefore, the roots of the equation are given as follows,

$x=\frac{-(8 a b) \pm \sqrt{16 a^{2} b^{2}}}{2\left(3 a^{2}\right)}$

$=\frac{-8 a b \pm 4 a b}{6 a^{2}}$

$=\frac{-4 b \pm 2 b}{3 a}$

Now we solve both cases for the two values of x. So, we have,

$x=\frac{-4 b+2 b}{3 a}$

$=-\frac{2 b}{3 a}$

Also,

$x=\frac{-4 b-2 b}{3 a}$

$=\frac{-2 b}{a}$

Therefore, the roots of the equation are $-\frac{2 b}{3 a}$ and $-\frac{2 b}{a}$.

(vii) We have been given, $3 x^{2}+2 \sqrt{5} x-5=0$

Now we also know that for an equation $a x^{2}+b x+c=0$, the discriminant is given by the following equation:

$D=b^{2}-4 a c$

Now, according to the equation given to us, we have, $a=3, b=2 \sqrt{5}$ and $c=-5$.

Therefore, the discriminant is given as,

$D=(2 \sqrt{5})^{2}-4(3)(-5)$

$=20+60$

 

$=80$

Since, in order for a quadratic equation to have real roots,.Here we find that the equation satisfies this condition, hence it has real roots.

Now, the roots of an equation is given by the following equation,

$x=\frac{-b \pm \sqrt{D}}{2 a}$

Therefore, the roots of the equation are given as follows,

$x=\frac{-(2 \sqrt{5}) \pm \sqrt{80}}{2(3)}$

$=\frac{-2 \sqrt{5} \pm 4 \sqrt{5}}{2(3)}$

$=\frac{-\sqrt{5} \pm 2 \sqrt{5}}{3}$

Now we solve both cases for the two values of x. So, we have

$x=\frac{-\sqrt{5}+2 \sqrt{5}}{3}$

$=\frac{\sqrt{5}}{3}$

Also,

$x=\frac{-\sqrt{5}-2 \sqrt{5}}{3}$

$=-\sqrt{5}$

Therefore, the roots of the equation are $\frac{\sqrt{5}}{3}$ and $-\sqrt{5}$.

(viii) We have been given, $x^{2}-2 x+1=0$

Now we also know that for an equation $a x^{2}+b x+c=0$, the discriminant is given by the following equation:

$D=b^{2}-4 a c$

Now, according to the equation given to us, we have, $a=1, b=-2$ and $c=1$.

 

Therefore, the discriminant is given as,

$D=(-2)^{2}-4(1)(1)$

$=4-4$

 

$=0$

Since, in order for a quadratic equation to have real roots,.Here we find that the equation satisfies this condition, hence it has real and equal roots.

Now, the roots of an equation is given by the following equation,

$x=\frac{-b \pm \sqrt{D}}{2 a}$

Therefore, the roots of the equation are given as follows,

$x=\frac{-(-2) \pm \sqrt{0}}{2(1)}$

$=\frac{2}{2}$

 

$=1$

Therefore, the roots of the equation are real and equal and its value is

(ix) We have been given, $2 x^{2}+5 \sqrt{3} x+6=0$

Now we also know that for an equation $a x^{2}+b x+c=0$, the discriminant is given by the following equation:

 

$D=b^{2}-4 a c$

Now, according to the equation given to us, we have, $a=2, b=5 \sqrt{3}$ and $c=6$.

 

Therefore, the discriminant is given as,

$D=(5 \sqrt{3})^{2}-4(2)(6)$

$=75-48$

 

$=27$

Since, in order for a quadratic equation to have real roots,.Here we find that the equation satisfies this condition, hence it has real roots.

Now, the roots of an equation is given by the following equation,

$x=\frac{-b \pm \sqrt{D}}{2 a}$

Therefore, the roots of the equation are given as follows,

$x=\frac{-(5 \sqrt{3}) \pm \sqrt{27}}{2(2)}$

$=\frac{-5 \sqrt{3} \pm 3 \sqrt{3}}{4}$

Now we solve both cases for the two values of x. So, we have,

$x=\frac{-5 \sqrt{3}+3 \sqrt{3}}{4}$

$=\frac{-\sqrt{3}}{2}$

Also,

$x=\frac{-5 \sqrt{3}-3 \sqrt{3}}{4}$

$=-2 \sqrt{3}$

Therefore, the roots of the equation are $-\frac{\sqrt{3}}{2} \mid$ and $-2 \sqrt{3}$.

(x) We have been given, $\sqrt{2} x^{2}+7 x+5 \sqrt{2}=0$

Now we also know that for an equation $a x^{2}+b x+c=0$, the discriminant is given by the following equation:

$D=b^{2}-4 a c$

Now, according to the equation given to us, we have, $a=\sqrt{2}, b=7$ and $c=5 \sqrt{2}$.

Therefore, the discriminant is given as,

$D=(7)^{2}-4(\sqrt{2})(5 \sqrt{2})$

$=49-40=9$

Since, in order for a quadratic equation to have real roots,.Here we find that the equation satisfies this condition, hence it has real roots.

Now, the roots of an equation is given by the following equation,

$x=\frac{-b \pm \sqrt{D}}{2 a}$

Therefore, the roots of the equation are given as follows,

$x=\frac{-(7) \pm \sqrt{9}}{2(\sqrt{2})}$

$=\frac{-7 \pm 3}{2 \sqrt{2}}$

Now we solve both cases for the two values of x. So, we have,

$x=\frac{-7+3}{2 \sqrt{2}}$

$=-\sqrt{2}$

Also,

$x=\frac{-7-3}{2 \sqrt{2}}$

$=-\frac{5}{\sqrt{2}}$

Therefore, the roots of the equation are $-\frac{5}{\sqrt{2}}$ and $-\sqrt{2}$

$D=(-2 \sqrt{2})^{2}-4(2)(1)$

$=8-8$

 

$=0$

Since, in order for a quadratic equation to have real roots,.Here we find that the equation satisfies this condition, hence it has real and equal roots.

Now, the roots of an equation is given by the following equation,

$x=\frac{-b \pm \sqrt{D}}{2 a}$

Therefore, the roots of the equation are given as follows,

$x=\frac{-(-2 \sqrt{2}) \pm \sqrt{0}}{2(2)}$

$=\frac{2 \sqrt{2}}{4}$

$=\frac{1}{\sqrt{2}}$

Therefore, the roots of the equation are real and equal and its value is $\frac{1}{\sqrt{2}}$

(xii) We have been given, $3 x^{2}-5 x+2=0$

Now we also know that for an equation $a x^{2}+b x+c=0$, the discriminant is given by the following equation:

$D=b^{2}-4 a c$

Now, according to the equation given to us, we have, $a=3, b=-5$ and $c=2$.

 

Therefore, the discriminant is given as,

$D=(-5)^{2}-4(3)(2)$

$=25-24$

 

$=1$

Since, in order for a quadratic equation to have real roots,.Here we find that the equation satisfies this condition, hence it has real roots.

Now, the roots of an equation is given by the following equation,

$x=\frac{-b \pm \sqrt{D}}{2 a}$

Therefore, the roots of the equation are given as follows,

$x=\frac{-(-5) \pm \sqrt{1}}{2(3)}$

$=\frac{5 \pm 1}{6}$

Now we solve both cases for the two values of x. So, we have,

$x=\frac{5+1}{6}$

$=1$

Also,

$x=\frac{5-1}{6}$

$=\frac{2}{3}$

Therefore, the roots of the equation are $\frac{2}{3}$ and $\square$.