Solve the given two equations and select the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For equation $I$: $8x^2 + 6x - 5 = 0$Factoring the quadratic equation: $8x^2 + 10x - 4x - 5 = 0$$2x(4x + 5) - 1(4x + 5) = 0$$(4x + 5)(2x - 1) = 0$

Vedclass · Accepted Answer

$x \leq y$

Vedclass · Answer

(C) For equation $I$: $8x^2 + 6x - 5 = 0$Factoring the quadratic equation: $8x^2 + 10x - 4x - 5 = 0$$2x(4x + 5) - 1(4x + 5) = 0$$(4x + 5)(2x - 1) = 0$Therefore,$x = -\frac{5}{4}$ or $x = \frac{1}{2}$.For equation $II$: $12y^2 - 22y + 8 = 0$Dividing by $2$: $6y^2 - 11y + 4 = 0$Factoring the quadratic equation: $6y^2 - 8y - 3y + 4 = 0$$2y(3y - 4) - 1(3y - 4) = 0$$(2y - 1)(3y - 4) = 0$Therefore,$y = \frac{1}{2}$ or $y = \frac{4}{3}$.Comparing the values:If $x = -1.25$,then $x If $x = 0.5$,then $x \leq y$ (since $y = 0.5$ or $1.33$).Combining these,we get $x \leq y$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $8x^2 + 6x = 5$
$II.$ $12y^2 - 22y + 8 = 0$

Similar Questions

If $x=z=225$ and $y=226$,then find the value of $x^{3}+y^{3}+z^{3}-3xyz$.

If $5x + 9y = 5$ and $125x^3 + 729y^3 = 120$,then the value of the product of $x$ and $y$ is:

Let $f(x) = x^2 - x + k - 2$,$k \in R$. Then the complete set of values of $k$ for which $y = |f(|x|)|$ is non-derivable at $5$ distinct points is:

If $\alpha$ and $\beta$ are the roots of the equation $6x^2 - 5x + 1 = 0$,then the value of $\tan^{-1} \alpha + \tan^{-1} \beta$ is:

If $x+y=2z$,find the value of $\frac{x}{x-z}+\frac{z}{y-z}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Solve the given two equations and select the correct answer from the given options.$I.$ $8x^2 + 6x = 5$$II.$ $12y^2 - 22y + 8 = 0$