First slide

Theory of expressions

Question

Let $f (x) = x^{2} + b_{1} x + c_{1}, g (x) = x^{2} + b_{2} x + c_{2}$ . Let the real roots of $f (x) = 0 be α, β$ and real roots of $g (x) = 0 be α + h, β + h$ . The least value of f(x) is -1/4. The least value of g(x) occurs at x = 7/2.

Moderate

Question

The least value of g(x) is

A

$- \frac{1}{4}$
B

-1
C

$- \frac{1}{3}$
D

$- \frac{1}{2}$

Solution

$\begin{array}{l} (β - α) = ((β + h) - (α + h)) \\ (β + α)^{2} - 4 αβ = [(β + h) + (α + h)]^{2} - 4 (β + h) (α + h) \\ {(- b_{1})}^{2} - 4 c_{1} = {(- b_{2})}^{2} - 4 c_{2} \\ D_{1} = D_{2} \end{array}$
The least value of f(x) is
$- \frac{D_{1}}{4} = - \frac{1}{4} \Rightarrow D_{1} = 1 and D_{2} = 1$
Therefore, the least value of
$g (x) is - \frac{D_{2}}{4} = - \frac{1}{4}$
The least value of g(x) occurs at
$- \frac{b_{2}}{2} = \frac{7}{2} or b_{2} = - 7$
Now, $b_{2}^{2} - 4 c_{2} = D_{2}$
or $49 - 4 c_{2} = 1 or \frac{48}{4} = c_{2} or c_{2} = 12$
$\Rightarrow x^{2} - 7 x + 12 = 0 or x = 3, 4$

Question

The value of b₂ is

A

-5
B

9
C

-8
D

-7

Solution

$\begin{array}{l} (β - α) = ((β + h) - (α + h)) \\ (β + α)^{2} - 4 αβ = [(β + h) + (α + h)]^{2} - 4 (β + h) (α + h) \\ {(- b_{1})}^{2} - 4 c_{1} = {(- b_{2})}^{2} - 4 c_{2} \\ D_{1} = D_{2} \end{array}$
The least value of f(x) is
$- \frac{D_{1}}{4} = - \frac{1}{4} \Rightarrow D_{1} = 1 and D_{2} = 1$
Therefore, the least value of
$g (x) is - \frac{D_{2}}{4} = - \frac{1}{4}$
The least value of g(r) occurs at
$- \frac{b_{2}}{2} = \frac{7}{2} or b_{2} = - 7$
Now, $b_{2}^{2} - 4 c_{2} = D_{2}$
or $49 - 4 c_{2} = 1 or \frac{48}{4} = c_{2} or c_{2} = 12$
$\Rightarrow x^{2} - 7 x + 12 = 0 or x = 3, 4$

Question

The roots of f(x) = 0 are

A

3, -4
B

-3, 4
C

3, 4
D

-3, -4

Solution

$\begin{array}{l} (β - α) = ((β + h) - (α + h)) \\ (β + α)^{2} - 4 αβ = [(β + h) + (α + h)]^{2} - 4 (β + h) (α + h) \\ {(- b_{1})}^{2} - 4 c_{1} = {(- b_{2})}^{2} - 4 c_{2} \\ D_{1} = D_{2} \end{array}$
The least value of f(x) is
$- \frac{D_{1}}{4} = - \frac{1}{4} \Rightarrow D_{1} = 1 and D_{2} = 1$
Therefore, the least value of
$g (x) is - \frac{D_{2}}{4} = - \frac{1}{4}$
The least value of g(r) occurs at
$- \frac{b_{2}}{2} = \frac{7}{2} or b_{2} = - 7$
Now, $b_{2}^{2} - 4 c_{2} = D_{2}$
or $49 - 4 c_{2} = 1 or \frac{48}{4} = c_{2} or c_{2} = 12$
$\Rightarrow x^{2} - 7 x + 12 = 0 or x = 3, 4$

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