(Each question 1 marks)

Question 11 Mark

Show the following quadratic equation by factorization method: $\text{x}^2+2\text{x}+5=0$

Answer

$x^2+2 x+5=0$ Now, completing the squares, we get $(x+1)^2+4=0 \Rightarrow(x+1)^2-2 i^2=0 \Rightarrow(x+1+2 i)(x+1-2 i)=$ $0 \Rightarrow(x+1+2 i)=0$ or $(x+1-2 i)=0 \therefore x=-1=2 i,-1+2 i$

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Question 21 Mark

If roots $\alpha,\beta$ of the equation $x^2− px + 16 = 0$ satisfy the relation $\alpha^2+\beta^2= 9,$ then write the value p.

Answer

$\alpha,\beta$ are the roots of the equation $x^2 - px + 16 = 0 \Rightarrow\alpha+\beta=\frac{-\text{b}}{\text{a}}=\frac{-(-\text{p})}{1}=\text{p}$ and $\alpha,\beta=\frac{\text{c}}{\text{a}}=\frac{16}{1}=16$ Now, $\alpha^2+\beta^2=9$ $\Rightarrow(\alpha+\beta)^2-2\alpha\beta=9$ $\Rightarrow\text{p}^2-2\times16=9$ $\Rightarrow\text{p}^2=9+32$ $\Rightarrow\text{P}=\sqrt{41}$

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Question 31 Mark

If $a$ and $b$ are roots of the equation $x^2-x+1=0$, then write the value of $a^2+b^2$.

Answer

The given equation is $x^2-x+1=0 \ldots$.(i) a and $b$ are the roots of (i) $\therefore a+b=-1 \ldots$ (ii) and $a b=1$ Now $a^2+b^2=(a+b) { }^2-2 \mathrm{ab}=(-1)^2-2.1=1-2=-1$

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Question 41 Mark

Show the following quadratic equation by factorization method: $9\text{x}^2+4=0$

Answer

$9 x^2+4=0 \Rightarrow(3 x)^2-\left(2 i^2\right)=0\left[\because i^2=-1\right] \Rightarrow(3 x+2 i)(3 x-2 i)=0 \Rightarrow\text{x}=\frac{-2}{3}\text{i}\ \text{or}\text{ x}=\frac{2}{3}\text{i}$ $\therefore\text{x}=\frac{-2}{3}\text{ i},\frac{2}{3}\text{ i}$

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Question 51 Mark

If $2+\sqrt{3}$ is root of the equation $x^2+ px + q = 0$, than write the values of p and q.

Answer

Since, a root of the equation $x^2+ px + q = 0$ .......(i) If $2+\sqrt{3}$ in one of the solution (roots) of in $2-\sqrt{3}$ will be other roots Now, sum of roots $=\frac{-\text{b}}{\text{a}}$ $\Rightarrow(2+\sqrt{3})+(2-\sqrt{3})=-\text{p}$ $\Rightarrow4=-\text{p}$ and product of root $=\frac{\text{c}}{\text{a}}$ $\Rightarrow(2+\sqrt{3})(2-\sqrt{3})=2$ ⇒ 4 - 3 = q ⇒ q = 1 Thus, p = -4, q = 1

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Question 61 Mark

If the difference between the roots of the equation is $x^2 + ax + 8 = 0$ is 2, write the values of a.

Answer

The given equation in $x^2 + ax + 8 = 0$ ....(i) Let $\alpha$ and $\beta$ are the two roots of (i) then $\alpha+\beta=\text{a}\ \&\ \alpha\beta=8$ we have given $\alpha -\beta=2$ Now $(\alpha-\beta)^2+4\alpha\beta=(\alpha+\beta)^2$ $\Rightarrow2^2+4.8=(\alpha+\beta)^2$ $\Rightarrow4+32=(\alpha+\beta)^2$ $\Rightarrow\alpha+\beta=\pm6$ But $\alpha+\beta=\text{a}$ $\therefore\alpha=\pm6$

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Question 71 Mark

Write roots of the equation $(a − b) x^2+ (b − c) x + (c − a) = 0$.

Answer

The given equation in $(a − b) x^2+ (b − c) x + (c − a) = 0$ ....(i) Let $\alpha$ and $\beta$ are the roots of (i) then, $\alpha+\beta=\frac{-\text{b}}{\text{a}}=\frac{-(\text{b}-\text{c})}{\text{a}-\text{b}}\ ...(\text{ii})$ and $\alpha\beta=\frac{\text{c}}{\text{a}}=\frac{\text{c}-\text{a}}{\text{a}-\text{b}}$ Now, $(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta$ $\Rightarrow\Big(-\frac{\text{b}-\text{c}}{\text{a}-\text{b}}\Big)^2-4\Big(\frac{\text{c}-\text{a}}{\text{a}-\text{b}}\Big)$ $\Rightarrow\frac{(\text{b}-\text{c})^2-4(\text{c}-\text{a})(\text{a}-\text{b})}{(\text{a}-\text{b})^2}$ $\therefore\alpha-\beta=\frac{\sqrt{(\text{b}-\text{c})^2-4(\text{c}-\text{a})(\text{a}-\text{b}})}{(\text{a}-\text{b})}\ ...(\text{ii})$ Solving (i) and (ii) $2\alpha=\frac{-(\text{b}-\text{c})}{\text{a}-\text{b}}+\frac{\sqrt{(\text{b}-\text{c})^2-4(\text{c}-\text{a})(\text{a}-\text{b}})}{\text{a}-\text{b}}$ $=\frac{-(\text{b}-\text{c})}{\text{a}-\text{b}}+\frac{\sqrt{(2\text{a}-\text{b}-\text{c})^2}}{\text{a}-\text{b}}$ $=\frac{2(\text{a}-\text{b})}{\text{a}-\text{b}}=2$ $\therefore\alpha=1$ From (ii) $\beta=\frac{\text{c}-\text{a}}{\text{a}-\text{b}}$

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Question 81 Mark

Write the number of real roots of the equation $(x-1)^2+(x-2)^2+(x-3)^2=0$

Answer

We have, $(x-1)^2+(x-2)^2+(x-3)^2=0$
$\Rightarrow x^2-2 x+1+x^2-4 x+4+x^2-6 x+9=0$
$\Rightarrow 3 x^2-12 x+14=0$
Now, $D=b^2-4 a c$
$=(-12)^2-4 \cdot 3 \cdot 14$
$=-24<0$
$\because D<0$ so, no real roots.

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Question 91 Mark

If $\alpha,\beta$ are roots of the equation $x^2− a (x + 1) − c = 0$, then write the value of $(1+\alpha)(1+\beta)$

Answer

$\alpha,\beta$ are the roots of $x^2 - a (x + 1) - c = 0 ...(i) \Rightarrow x^2 - ax - (a + c) = 0 \alpha+\beta=\frac{-\text{b}}{\text{a}}=\text{a}$ and $\alpha\beta=\frac{\text{c}}{\text{a}}=-(\text{a}+\text{c})\ ...(\text{ii})$ Now, $(1+\alpha)(1+\beta)=1+(\alpha+\beta)+\alpha\beta$ $=1+\text{a}+(-\text{a}-\text{c})$ $=1-\text{c}$

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Question 101 Mark

If $\alpha,\beta$ are roots of the equation $x^2 + lm + m = 0$, write an equation whose roots are $-\frac{1}{\alpha}$ and $-\frac{1}{\beta}$

Answer

$\alpha$ and $\beta$ are the roots of $x^2 + lm + m = 0$ ...(i) $\therefore\alpha+\beta=\frac{-\text{b}}{\text{a}}=-\text{l}$ and $\alpha\beta=\frac{\text{c}}{\text{a}}=\text{m}$ now, The quadratic equation whose roots are $-\frac{1}{\alpha}$ and $-\frac{1}{\beta}$ is $x^2$ - (sum of roots)x + (product of roots) = 0 $\Rightarrow\text{x}^2-\Big(-\frac{1}{\alpha}+\frac{-1}{\beta}\Big)+\Big(-\frac{1}{\alpha}.\frac{-1}{\beta}\Big)=0$ $\Rightarrow\text{x}^2+\Big(\frac{\alpha+\beta}{\alpha\beta}\Big)\text{x}+\frac{1}{\alpha\beta}=0$ $\Rightarrow\text{x}^2+\Big(\frac{-\text{I}}{\text{m}}\Big)\text{x}+\frac{1}{\text{m}}=0$ $\Rightarrow\text{mx}^2-\text{lx}+1=0$

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Question 111 Mark

Write the number of quadratic equations, with real roots, which do not change by squaring their roots.

Answer

Let $a$ and $b$ be the real roots of the quadratic equation. we need to find the number of quadratic equation such that they ramain unchanged even if roots are squared. $a^2=a$ and $b^2=b \Rightarrow a(a-1)=0$ and $b(b-1)=0 \Rightarrow a=0$ or $a=1$ and $b=0$ or $b=1$ so we have four pairs of roots $(0,0),(0,1),(1,0),(1,1)$ For $(0,0)(x-0)(x-0)=x^2$ for $(0,1)(x-0)$ $(x-1)=x(x-1)=x^2-1$ For $(1,0)(x-1)(x-0)=(x-1) x=x^2-1$ For $(1,1)(x-1)(x-1)=(x-1)^2=x^2-2 x+1$ So there are 3 quadratic equations with real roots, which do not change by squaring their roots.

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Question 121 Mark

If a and b are roots of the equation $x^2− px + q = 0$, than write the value of $\frac{1}{\text{a}}+\frac{1}{\text{b}}.$

Answer

Since, a, b are the roots of $x^2 - px + q = 0 \Rightarrow\text{a}+\text{b}=-\Big(\frac{-\text{P}}{1}\Big)=\text{P}$ $\text{a.b}=\frac{\text{q}}{1}=\text{q}$ Now, $\frac{1}{\text{a}}+\frac{1}{\text{b}}=\frac{\text{a}+\text{b}}{\text{a.b}}=\frac{\text{P}}{\text{q}}$ $\therefore\frac{1}{\text{a}}+\frac{1}{\text{b}}=\frac{\text{p}}{\text{q}}$

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Question 131 Mark

Show the following quadratic equation by factorization method: $x^2 + 1 = 0$

Answer

$x^2+1=0 \Rightarrow x^2+i^2=0\left[\because i^2=-1\right] \Rightarrow(x+i)(x-i)=0\left[a^2-b^2=(a+b)(a-b)\right] \Rightarrow x=i,-i$

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