mathematics-class-11-unit-05-chapter-01-problem-solving-quadratic-equations-l-1

<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Problems on quadratic equation lecture - 1</B></Success></p>
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<p><strong>1)</strong> For $ax^2 + bx + c = 0$</p>
<p>The point of maxima or minima are attained at $ x = -\frac{b}{2a}$</p>
<p><strong>2)</strong> $\alpha$ & $\beta$ roots of $a x^2+b x+c=0$</p>
<p>$\alpha+\beta=-b/a$</p>
<p>$\alpha \beta=c/a$</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Problems on quadratic equation lecture - 1</B></Success></p>
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<p>$ x^2+x-3=0 $</p>
<p>$ x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$</p>
<p>If $\alpha$ & $\beta$ are roots of the given equation.</p>
<p>$\alpha^3-4 \beta^2+19=\text { ? }$</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Problems on quadratic equation lecture - 1</B></Success></p>
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<p>$(x-a)(x-a-b)=1; a, b \in \R $</p>
<p>Prove that there are 2 real roots $\alpha$ & $\beta$
such that $\alpha< a<\beta$.</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Problems on quadratic equation lecture - 1</B></Success></p>
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<p>Put $x-a=y$</p>
<p>$y(y-b)=1 $</p>
<p>$y^2-b y-1=0$</p>
<p>$D=b^2+4>0$</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Problems on quadratic equation lecture - 1</B></Success></p>
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<p>Thus are 2 distinct real roots, say $\alpha $ & $\beta$ :</p>
<p>$\alpha \beta=-1 \Rightarrow \alpha $ &  $ \beta$ have opp. signs</p>
<p>$ \alpha<0<\beta $</p>
<p>$ x=a+y $</p>
<p>$ \alpha<a<\beta $</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Problems on quadratic equation lecture - 1</B></Success></p>
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<p>$2 x^2-3 x-1=0 \leftarrow p \text { is a root} $</p>
<p>$ x^2+3 x-2=0 \leftarrow q \text { is a root }$</p>
<p>Given $p q \neq 1$. Find the value of $\frac{p q+p+1}{q}$</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Problems on quadratic equation lecture - 1</B></Success></p>
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<p>$ 2 p^2-3 p-1=0 \longrightarrow ~ (1)$</p>
<p>$ q^2+3 q-2=0 \longrightarrow ~ (2)$</p>
<p>$-q^2\left[-1 - \frac{3}{q}+\frac{2}{q^2}\right]=0 $</p>
<p>$\Rightarrow 2 \cdot \frac{1}{q^2}-\frac{3}{q}-1=0$</p>
<p>$\frac{1}{q}$ is a root of the equation $2x^2-3x-1=0$</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Problems on quadratic equation lecture - 1</B></Success></p>
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<p>$\therefore p+\frac{1}{q}=\frac{3}{2}$  and  $\frac {p}{q}=-\frac{1}{2}.$</p>
<p>$\frac{p q+p+1}{q} =p+\frac{p}{q}+\frac{1}{q} $</p>
<p>$ =3 / 2-1 / 2=1$</p>
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<p>$a x^2+b x+c=0 \longleftarrow \alpha$ is a root</p>
<p>$-a x^2+b x+c=0 \longleftarrow \beta$ is a root.</p>
<p>Show: $ ~ \frac{a}{2} x^2+b x+c=0$ has a root between $\alpha ~ \& ~ \beta$.</p>
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<p>Take $p(x)=\frac{a}{2} x^2+b x+c$</p>
<p>$ p(\alpha)=\frac{a}{2} \alpha^2+b \alpha+c=\left[a \alpha^2+b \alpha+c\right]-\frac{a}{2} \alpha^2=-\frac{a}{2} \alpha^2 $</p>
<p>$ p(\beta)=\frac{a}{2} \beta^2+b \beta+c=\left[-a \beta^2+b \beta+c\right]+\frac{3 a}{2} \beta^2=\frac{3 a}{2} \beta^2$</p>
<p>$\therefore p(\alpha)  p(\beta)$ have opposite Signs.</p>
<p>$p(x)=0$ for some $x \in(\alpha, \beta)$.</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Problems on quadratic equation lecture - 1</B></Success></p>
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<p>Case (i) $ ~ |x-2|^2+|x-2|-2=0$ Verify that Roots are 3 & -1.</p>
<p>Case (ii) $ ~ $ $\left|x^2+4 x+3\right|+2 x+5=0$.</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Problems on quadratic equation lecture - 1</B></Success></p>
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<p>$ x^2+4 x+3>0 $</p>
<p>$ (x+3)(x+1)>0 $</p>
<p>$ x<-3 \text { or } x>-1 .$</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Problems on quadratic equation lecture - 1</B></Success></p>
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<p>Given equation becomes $x^2+4 x+3+2 x+5=0$</p>
<p>(ie) $x^2+6 x+8=0$</p>
<p>$(x + 4) (x + 2) =0 $</p>
<p>$\Rightarrow x =  -4 ~ $ &$ ~ -2$</p>
<p>$x = -4$ is the only solution.</p>
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