If 2 x2-6 x-4-4 x2-10 x-3-x2-3 x-2 x2-5 x- value of x is for x-0-

The correct option is A 115We have nbsp;2x^2 6x +4/4x^2 10x +3 = nbsp;x^2 3x/2x^2 5x . If we multiply numerator and denominator of the RHS by 2, we will ge ...

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Byju's Answer

Standard IX

Mathematics

Equal Ratio

If 2 x2-6 x+4...

Question

If $\frac{2 x^{2} - 6 x + 4}{4 x^{2} - 10 x + 3}$ = $\frac{x^{2} - 3 x}{2 x^{2} - 5 x}$ , value of x is for $x \neq$ 0.

\frac{11}{5}

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\frac{5}{11}

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Solution

The correct option is A $\frac{11}{5}$
We have $\frac{2 x^{2} - 6 x + 4}{4 x^{2} - 10 x + 3}$ = $\frac{x^{2} - 3 x}{2 x^{2} - 5 x}$ .

If we multiply numerator and denominator of the RHS by 2, we will get
$\frac{2 x^{2} - 6 x + 4}{4 x^{2} - 10 x + 3}$ = $\frac{2 x^{2} - 6 x}{4 x^{2} - 10 x}$ .

As per the theorem of equivalent ratio,
$⟹$ $\frac{a}{b}$ = $\frac{c}{d}$ = $\frac{a - c}{b - d}$

Hence,
$\frac{2 x^{2} - 6 x + 4}{4 x^{2} - 10 x + 3}$ = $\frac{2 x^{2} - 6 x}{4 x^{2} - 10 x}$

= $\frac{2 x^{2} - 6 x + 4 - (2 x^{2} - 6 x)}{4 x^{2} - 10 x + 3 - (4 x^{2} - 10 x)}$
= $\frac{4}{3}$

So we got the ratio equals to $\frac{4}{3}$ and it is equal to the $\frac{x^{2} - 3 x}{2 x^{2} - 5 x}$ .

Hence, $\frac{x^{2} - 3 x}{2 x^{2} - 5 x}$ = $\frac{4}{3}$

$3 x^{2} - 9 x = 8 x^{2} - 20 x$
$5 x^{2} - 11 x = 0$
$x (5 x - 11) = 0$

$x = 0, \frac{11}{5}$

$x = \frac{11}{5} a s x \neq 0 (g i v e n)$

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Similar questions

Dividing $4 x^{2} - 10 x + 6$ by $2$ , we get ______.

Now for each pair of polynomials p(x) and q(x) given below, find p(x) + q(x) and p(x) − q(x). Also compute p(1) + q(1) and p(1) − q(1) for each.

(i)

(ii)

(iii)

(iv)

(v)

Q. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
(i)

2 x^{2} - 7 x + 3 = 0

(ii)

2 x^{2} + x - 4 = 0

(iii)

4 x^{2} + 4 \sqrt{3 x} + 3 = 0

(iv)

2 x^{2} + x + 4 = 0

Q. Evaluate the following:

\begin{matrix} {lim}_{x \to 1} \frac{x^{2} - 3 x + 2}{x^{2} - 4 x + 3} {lim}_{x \to 2} \frac{x^{3} + x^{2} - 12}{x^{3} - x^{2} - x - 2} {lim}_{x \to \sqrt{2}} \frac{x^{2} + x \sqrt{2} - 4}{x^{2} - 3 x \sqrt{2} + 4} {lim}_{x \to 3} \frac{x^{3} - 3 x^{2} - x + 3}{x^{3} - 6 x^{2} + 9 x} \end{matrix}

Q. Find the values of a, b, c for following quadratic equations by comparing with standard form.

(i) x² + 2x + 1 = 0
(ii) 2x² – x + 3 = 0
(iii) x² – x – 3 = 0
(iv) x² + 5x – 4 = 0
(v) x² – 7x + 4 = 0
(vi) 4x² – 9x + 1 = 0

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If 2x2−6x+44x2−10x+3 = x2−3x2x2−5x , value of x is for x≠0.

If $\frac{2 x^{2} - 6 x + 4}{4 x^{2} - 10 x + 3}$ = $\frac{x^{2} - 3 x}{2 x^{2} - 5 x}$ , value of x is for $x \neq$ 0.