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

(B) For equation $I$: $2x^2 + 13x - 7 = 0$Using the quadratic formula or factorization: $2x^2 + 14x - x - 7 = 0$$2x(x + 7) - 1(x + 7) = 0$$(2x - 1)(x

Vedclass · Accepted Answer

if $x < y$

Vedclass · Answer

(B) For equation $I$: $2x^2 + 13x - 7 = 0$Using the quadratic formula or factorization: $2x^2 + 14x - x - 7 = 0$$2x(x + 7) - 1(x + 7) = 0$$(2x - 1)(x + 7) = 0$So,$x = 0.5$ or $x = -7$.For equation $II$: $2y^2 - 5y + 3 = 0$Using the quadratic formula or factorization: $2y^2 - 2y - 3y + 3 = 0$$2y(y - 1) - 3(y - 1) = 0$$(2y - 3)(y - 1) = 0$So,$y = 1.5$ or $y = 1$.Comparing the values:When $x = 0.5$,$x When $x = -7$,$x In all cases,$x < y$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $2x^2 + 13x - 7 = 0$
$II.$ $2y^2 - 5y + 3 = 0$

Similar Questions

If $2 + 3i$ is one of the roots of the equation $2x^3 - 9x^2 + kx - 13 = 0,$ where $k \in R,$ then the real root of this equation

Solve the given two equations and select the correct answer from the given options.
$I. 3x^2 - 4x - 32 = 0$
$II. 2y^2 - 17y + 36 = 0$

If the ratio of the roots of $x^2 + bx + c = 0$ and $x^2 + qx + r = 0$ is the same,then:

If the graph of $y = ax^3 + bx^2 + cx + d$ is symmetric about the line $x = k$,then:

Solve the given two equations and select the correct answer from the given options.
$I.$ $16x^2 + 20x + 6 = 0$
$II.$ $10y^2 + 38y + 24 = 0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Solve the given two equations and select the correct answer from the given options.$I.$ $2x^2 + 13x - 7 = 0$$II.$ $2y^2 - 5y + 3 = 0$