1 Marks Question

Question 11 Mark

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

Answer

x² + 2x + 5 = 0
Now, completing the squares, we get
(x + 1)² + 4 = 0
⇒ (x + 1)² - 2i² = 0
⇒ (x + 1 + 2i) (x + 1 - 2i) = 0
⇒ (x + 1 + 2i) = 0 or (x + 1 - 2i) = 0
$\therefore$ x = -1 = 2i, -1 + 2i

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Question 21 Mark

If roots $\alpha,\beta$ of the equation x²− 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 x² - 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 x²− x + 1 = 0, then write the value of a² + b².

Answer

The given equation is x²− x + 1 = 0 ....(i)
a and b are the roots of (i)
$\therefore$ a + b = -1 ...(ii)
and ab = 1
Now a² + b² = (a + b)² - 2ab
= (-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

9x² + 4 = 0
⇒ (3x)² - (2i²) = 0 [$\because$ i² = -1]
⇒ (3x + 2i) (3x - 2i) = 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 $\alpha,\beta$ are roots of the equation x²− a (x + 1) − c = 0, then write the value of $(1+\alpha)(1+\beta)$

Answer

$\alpha,\beta$ are the roots of x² - a (x + 1) - c = 0 ...(i)
⇒ x² - 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 61 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.
a² = a and b² = b
⇒ a(a - 1) = 0 and b(b - 1) = 0
⇒ 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) = x²
for (0, 1)
(x - 0) (x - 1) = x (x - 1) = x² - 1
For (1, 0)
(x - 1) (x - 0) = (x - 1) x = x² - 1
For (1, 1)
(x - 1) (x - 1) = (x - 1)² = x²- 2x + 1
So there are 3 quadratic equations with real roots, which do not change by squaring their roots.

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Question 71 Mark

If $\alpha,\beta$ are roots of the equation x² + lm + m = 0, write an equation whose roots are $-\frac{1}{\alpha}$ and $-\frac{1}{\beta}$

Answer

$\alpha$ and $\beta$ are the roots of x² + 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 x² - (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 81 Mark

If a and b are roots of the equation x²− px + q = 0, than write the value of $\frac{1}{\text{a}}+\frac{1}{\text{b}}.$

Answer

Since,
a, b are the roots of x² - 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 91 Mark

If $2+\sqrt{3}$ is root of the equation x²+ px + q = 0, than write the values of p and q.

Answer

Since,
a root of the equation
x²+ 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 101 Mark

Write roots of the equation (a − b) x²+ (b − c) x + (c − a) = 0.

Answer

The given equation in (a − b) x²+ (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 111 Mark

Show the following quadratic equation by factorization method:
x² + 1 = 0

Answer

x² + 1 = 0
⇒ x²+ i² = 0 [$\because$ i² = -1]
⇒ (x + i) (x - i) = 0 [a² - b² = (a + b) (a - b)]
⇒ x = i, -i

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Question 121 Mark

If the difference between the roots of the equation is x² + ax + 8 = 0 is 2, write the values of a.

Answer

The given equation in x² + 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 131 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

⇒ x² - 2x + 1 + x²- 4x + 4 + x² - 6x + 9 = 0

⇒ 3x² - 12x + 14 = 0

Now, D = b² - 4ac

= (-12)² - 4.3.14

= -24 < 0

$\because$ D < 0 so, no real roots.

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