1 Marks Question

Question 11 Mark

State whether the following quadratic equations have two distinct real roots. Justify your answer.
$\big(\text{x}-\sqrt{2}\big)^2-2(\text{x}+1)=0.$

Answer

Main concept used:
Quadratic equation $a x^2+b x+c=0$ will have two distinct real roots if $D >0$ or $b ^2-4 ac >0$.
$(x-\sqrt{2})^2-2(x+1)=0$
$\Rightarrow x^2-2 \sqrt{2} x+2-2 x-2=0$
$\Rightarrow x^2-(2 \sqrt{2}+2) x=0$
Now, $D=b^2-4 a c$
$=[-(2 \sqrt{2}+2)]^2-4 \times 1 \times 0$
${[\therefore a=1, b=-(2 \sqrt{2}+2), c=0]}$
$=(2 \sqrt{2}+2)^2>0$
As $D>0$, so the given equation has real and unequal roots.

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Question 21 Mark

Does there exist a quadratic equation whose coefficients are all distinct irrationals but both the roots are rationals? Why?

Answer

Yes, there may be a quadratic equation whose coefficients are all distinct irrationals, but both the roots are rational.
For example, consider a quadratic equation having distinct irrational coefficients
$3\sqrt{\frac{3}{2}}\text{x}^2+\frac{5}{\sqrt{6}}\text{x}-2\sqrt{\frac{2}{3}}=0$
Now $D = b^2 - 4ac$
$=\bigg[\frac{5}{\sqrt{6}}\bigg]^2-4\bigg[\frac{3\sqrt{3}}{2}\bigg]\bigg[\frac{-2\sqrt{2}}{3}\bigg]$
$\bigg(\text{a}=\frac{3\sqrt{3}}{2},\text{ b}=\frac{5}{\sqrt{6}},\text{ c}=\frac{-2\sqrt{2}}{\sqrt{3}}\bigg)$
$=\frac{25}{6}+\frac{24}{1}=\frac{25+144}{6}$
$\Rightarrow\ \text{D}=\frac{169}{6}\Rightarrow\ \sqrt{\text{D}}=\frac{13}{\sqrt{6}}$
Roots are given by $\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}=\frac{-\frac{5}{\sqrt{6}}+\frac{13}{\sqrt{6}}}{2.\frac{3\sqrt{3}}{2}}$
$=\frac{\frac{1}{\sqrt{6}}[-5\pm13]\sqrt{2}}{6\sqrt{3}}=\frac{(-5\pm13)\sqrt{2}}{\sqrt{2}\sqrt{3}\times6\sqrt{3}}$
$\Rightarrow\ \text{x}=\frac{(-5\pm13)}{18}$
$\Rightarrow\ \text{x}=\frac{8}{18}$ or $\frac{-18}{18}$
$\Rightarrow\ \text{x}=\frac{4}{9}$ or -1
Hence, the roots are rational while coefficeints a, b, c were irrational.

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