I. 6x^2 + 41x + 63 = 0II. 4y^2 + 8y + 3 = 0 | vedclass

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

Question

(C) For equation $I$: $6x^2 + 41x + 63 = 0$$6x^2 + 27x + 14x + 63 = 0$$3x(2x + 9) + 7(2x + 9) = 0$$(3x + 7)(2x + 9) = 0$$x = -7/3 \approx -2.33$ or $x

Vedclass · Accepted Answer

$x < y$

Vedclass · Answer

(C) For equation $I$: $6x^2 + 41x + 63 = 0$$6x^2 + 27x + 14x + 63 = 0$$3x(2x + 9) + 7(2x + 9) = 0$$(3x + 7)(2x + 9) = 0$$x = -7/3 \approx -2.33$ or $x = -9/2 = -4.5$For equation $II$: $4y^2 + 8y + 3 = 0$$4y^2 + 6y + 2y + 3 = 0$$2y(2y + 3) + 1(2y + 3) = 0$$(2y + 1)(2y + 3) = 0$$y = -1/2 = -0.5$ or $y = -3/2 = -1.5$Comparing the values:$x = -2.33, -4.5$$y = -0.5, -1.5$Since all values of $x$ are less than all values of $y$,we conclude that $x < y$.

$I. 6x^2 + 41x + 63 = 0$
$II. 4y^2 + 8y + 3 = 0$

Similar Questions

If $ax^2 + bx + c = 0$ and $bx^2 + cx + a = 0$ have a common root and $a, b, c$ are non-zero real numbers,then $\frac{a^3 + b^3 + c^3}{abc} = $

The length of a rectangular plot is $8 \, m$ greater than its breadth. If the area of the plot is $308 \, m^{2}$,find the length of the plot.

If $a + b + c = 0$,$a \neq 0$,$a, b, c \in Q$,then both the roots of the equation $ax^2 + bx + c = 0$ are

Solve the given two equations and select the correct answer from the given options.
$I.$ $18x^2 - 13\sqrt{7}x + 14 = 0$
$II.$ $32y^2 - 19\sqrt{6}y + 9 = 0$

Which of the following equations has equal roots?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$I. 6x^2 + 41x + 63 = 0$$II. 4y^2 + 8y + 3 = 0$