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State whether the following quadratic equations have two distinct real roots. Justify your answer.
\( x^{2}-3 x+4=0 \)
To do:
We have to state whether the given quadratic equations have two distinct real roots.
Solution:
(i) $x^{2}-3 x+4=0$
Comparing with $a x^{2}+b x+c=0$, we get,
$a =1, b=-3$ and $c=4$
Discriminant $D=b^{2}-4 a c$
$=(-3)^{2}-4(1)(4)$
$=9-16$
$=-7<0$
$D<0$
Hence, the equation $x^{2}-3 x+4=0$ has no real roots.
(ii) $2 x^{2}+x-1=0$
Comparing with $a x^{2}+b x+c=0$, we get,
$a=2, b=1$ and $c=-1$
Discriminant $D=b^{2}-4 a c$
$=(1)^{2}-4(2)(-1)$
$=1+8$
$=9>0$
$D>0$
Hence, the equation $2 x^{2}+x-1=0$ has two distinct real roots.
(iii) $2 x^{2}-6 x+\frac{9}{2}=0$
Comparing with $a x^{2}+b x+c=0$, we get,
$a =2, b=-6$ and $c=\frac{9}{2}$
Discriminant $D=b^{2}-4 a c$
$=(-6)^{2}-4(2)(\frac{9}{2})$
$=36-36$
$=0$
$D=0$
Hence, the equation $2 x^{2}-6 x+\frac{9}{2}=0$ has equal and real roots.
(iv) $3 x^{2}-4 x+1=0$
Comparing with $a x^{2}+b x+c=0$, we get,
$a =3, b=-4$ and $c=1$
Discriminant $D=b^{2}-4 a c$
$=(-4)^{2}-4(3)(1)$
$=16-12$
$=4>0$
$D>0$
Hence, the equation $3x^{2}-4 x+1=0$ has two distinct real roots.
(v) \( (x+4)^{2}-8 x=0 \)
$x^2+4^2+2(4)x-8x=0$
$x^2+8x-8x+16=0$
$x^2+16=0$
Comparing with $a x^{2}+b x+c=0$, we get,
$a =1, b=0$ and $c=16$
Discriminant $D=b^{2}-4 a c$
$=(0)^{2}-4(1)(16)$
$=0-64$
$=-64<0$
$D<0$
Hence, the equation \( (x+4)^{2}-8 x=0 \) has no real roots.
(vi) \( (x-\sqrt{2})^{2}-2(x+1)=0 \)
$x^2+(\sqrt2)^2-2(\sqrt2)x-2x-2=0$
$x^2+2-2\sqrt2x-2x-2=0$
$x^2-(2\sqrt2+2)x=0$
Comparing with $a x^{2}+b x+c=0$, we get,
$a =1, b=2(\sqrt2+1)$ and $c=0$
Discriminant $D=b^{2}-4 a c$
$=[2(\sqrt2+1)]^{2}-4(1)(0)$
$=4(\sqrt2+1)^2-4$
$=4[(\sqrt2+1)^2-1]>0$
$D>0$
Hence, the equation \( (x-\sqrt{2})^{2}-2(x+1)=0 \) has two distinct real roots.
(vii) \( \sqrt{2} x^{2}-\frac{3}{\sqrt{2}} x+\frac{1}{\sqrt{2}}=0 \)
Comparing with $a x^{2}+b x+c=0$, we get,
$a=\sqrt{2}, b=-\frac{3}{\sqrt{2}}$ and $c=\frac{1}{\sqrt{2}}$
Therefore,
Discriminant $D=b^{2}-4 a c$
$=(-\frac{3}{\sqrt{2}})^{2}-4 \sqrt{2}(\frac{1}{\sqrt{2}})$
$=\frac{9}{2}-4$
$=\frac{9-8}{2}$
$=\frac{1}{2}>0$
$D>0$
Hence, the equation \( \sqrt{2} x^{2}-\frac{3}{\sqrt{2}} x+\frac{1}{\sqrt{2}}=0 \) has two distinct real roots.
(viii) $x(1-x)-2=0$
$x^{2}-x+2=0$
Comparing with $a x^{2}+b x+c=0$, we get,
$a=1, b=-1$ and $c=2$
Therefore,
Discriminant $D=b^{2}-4 a c$
$=(-1)^{2}-4(1)(2)$
$=1-8$
$=-7<0$
$D<0$
Hence, the equation $x(1-x)-2=0$ has no real roots.
(ix) $(x-1)(x+2)+2=0$
$x^{2}+x-2+2=0$
$x^{2}+x+0=0$
Comparing the above equation with $a x^{2}+b x+c-0$, we get,
$a=1, b=1$ and $c=0$
Therefore,
Discriminant $D=b^{2}-4 a c$
$=1-4(1)(0)$
$=1>0$
$D>0$
Hence, the equation $(x-1)(x+2)+2=0$ has two distinct real roots.
(x) $(x+1)(x-2)+x=0$
$x^{2}+x-2 x-2+x=0$
$x^{2}-2=0$
Comparing with $a x^{2}+b x+c=0$, we get,
$a=1, b=0$ and $c=-2$
Therefore,
Discriminant $D=b^{2}-4 a c$
$=(0)^{2}-4(1)(-2)$
$=0+8$
$=8>0$
Hence, the equation $(x+1)(x-2)+x=0$ has two distinct real roots.