# State whether the following quadratic equations have two distinct real roots. Justify your answer.$x^{2}-3 x+4=0$

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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.

Updated on 10-Oct-2022 13:27:26

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