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3 years ago

If(a²-1) x²+(a-1)x+a²-4a+3=0 is an identity in x, then find the value of a

Mathematics

1 answer:

PtichkaEL [24]3 years ago

6 0

Answer:

Step-by-step explanation:

$(a^2-1)x^2+(a-1)x+a^2-4a+3=0\\\\Calculate\ and\ identify\ the\ polynomials\\\\\Longleftrightarrow\ a^2x^2-x^2+ax-x+a^2-4a+3=0\\\\\Longleftrightarrow\ a^2x^2+ax+a^2-4a+3=x^2+x+0\\\\\Longleftrightarrow\ \left\{\begin{array}{ccc}a^2&=&1\\a&=&1\\a^2-4a+3&=&0\\\end{array} \right.\\\\\Longleftrightarrow\ \left\{\begin{array}{ccc}(a-1)(a+1)&=&0\\a-1&=&0\\(a-1)(a-3)&=&0\\\end{array} \right.\\\\\\We\ must\ exclude\ a=-1\ and\ a=3\ (not\ solution)\\\Longrightarrow\ a=1\\$

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3 years ago

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Answer:

A. The football does not reach a height of 15ft

Step-by-step explanation:

Given

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Required

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The option illustrates the height reached by the ball.

To solve this, we make use of maximum of a function

For a function f(x)

Such that:

$f(x) = ax^2 + bx + c$ :

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i.e we first solve for $\frac{-b}{2a}$

Then substitute $\frac{-b}{2a}$ for x in $f(x) = ax^2 + bx + c$

In our case:

First we need to solve $\frac{-b}{2a}$

Then substitute $\frac{-b}{2a}$ for t in $h(t) = -16t^2 + 30t$

By comparison:

$b = 30$

$c = -16$

$\frac{-b}{2a} = \frac{-30}{2 * -16}$

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$\frac{-b}{2a} = \frac{30}{32}$

$\frac{-b}{2a} = \frac{15}{16}$

Substitute $\frac{15}{16}$ for t in $h(t) = -16t^2 + 30t$

$h(\frac{15}{16}) = -16(\frac{15}{16})^2 + 30(\frac{15}{16})$

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$h(\frac{15}{16}) = -(\frac{225}{16}) + \frac{450}{16}$

$h(\frac{15}{16}) = \frac{-225 + 450}{16}$

$h(\frac{15}{16}) = \frac{225}{16}$

$h(\frac{15}{16}) = 14.0625$

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4 years ago

Read 2 more answers

Can someone help me?

Wittaler [7]

Answer:

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Step-by-step explanation:

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3 years ago

If(a²-1) x²+(a-1)x+a²-4a+3=0 is an identity in x, then find the value of a​

If(a²-1) x²+(a-1)x+a²-4a+3=0 is an identity in x, then find the value of a